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X 


THE 

ELEMENTS 

OF 

GEOMETRY    X 

BUSH  AND  CLARKE 


SILWR^BURDETT  &  COMPANY 
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MUWIHWWtlH 


IN  MEMORIAM 
FLORIAN  CAJORI 


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Digitized  by  the  Internet  Arciiive 

in  2008  with  funding  from 

IVIicrosoft  Corporation 


http://www.archive.org/details/elementsofgeometOObushrich 


THE 


ELEMENTS  OF  GEOMETRY 


BY 


WALTER   N.    BUSH 

PRINCIPAL   OF  THE   POLYTECHNIC   HIGH   SCHOOL 
SAN   FRANCISCO 

AND 

JOHN  B.    CLARKE 

DEPARTMENT   OF  MATHEMATICS,   POLYTECHNIC   HIGH  SCHOOL 
SAN  FRANCISCO 


SILVER,   BURDETT   AND   COMPANY 

NEW  YORK         BOSTON         CHICAGO         SAN  FRANCISCO 


Copyright,  1905, 1909, 

Bt  silver,  burdett  and  company. 


CAJORI 


PREFACE 

The  present  text,  the  outgrowth  of  over  twenty  years'  expe- 
rience with  classes  in  geometry,  carries  the  work  from  the 
simple  elements  necessary  in  the  beginning  class  of  the  high 
school  to  the  most  advanced  requirements  of  university  prepa- 
ration. For  many  years  we  have  followed  in  our  own  classes 
the  plan  here  presented,  with  such  modifications  and  improve- 
ments as  experience  has  suggested. 

It  is  generally  conceded  that  much  of  the  pupil's  difficulty 
in  demonstration  arises  from  his  failure  to  grasp  thoroughly 
and  keep  vividly  in  mind  as  separate  and  distinct  statements, 
first,  the  exact  data  of  the  proposition,  and,  second,  the  precise 
fact  to  be  established.  To  remove  this  stumbling-block,  we 
have  stated  the  hypothesis  and  conclusion  separately  for  every 
theorem  and  corollary  demonstrated. 

Through  the  "  open  "  arrangement  of  the  printed  matter,  we 
have  sought  to  make  each  successive  step  stand  out  clearly ; 
and  by  so  adjusting  diagrams  and  text  that  in  the  course  of 
any  single  demonstration  it  is  unnecessary  to  turn  the  page, 
we  have  endeavored  to  avoid  waste  of  effort  on  the  part  of 
the  pupil. 

All  the  original  exercises  should  be  mastered  with  only  such 
help  as  is  given  by  the  book  itself.  If  a  proposition  has  been 
found  difficult  for  the  average  pupil,  it  has  been  broken  up 
into  a  series  of  exercises  in  such  sequence  that  the  difficulties 
are  presented  one  at  a  time  and  in  natural  order,  the  truth 


iV  PREFACE 

of  tht  main  propoiition  being  •stablished  by  means  of  these 
graded  exercises. 

It  has  been  our  purpose  to  eliminate  discouraging  elements, 
to  refresh  the  memory  of  the  student  before  he  begins  inven- 
tive work,  to  arouse  his  interest  and  to  inspire  his  confidence 
in  his  ability  to  discover  hidden  truths. 

We  desire  to  express  our  acknowledgments  for  valuable 
counsel  and  suggestions  to  Professor  F.  N.  Cole  of  Columbia 
University,  to  Professor  Irving  Stringham  of  the  University 
of  California,  to  Professor  R.  E.  Gaines  of  Richmond  College, 
Virginia,  and  to  J.  A.  C.  Chandler,  LL.D.,  formerly  Dean  of 
Richmond  Academy,  Virginia. 

WALTER  N.  BUSH. 
JOHN  B.  CLARKE. 

San  Francisco, 
Caxifoknia. 


PLAN  AND  SCOPE 

There  are  few  students  who  fail  to  respond  to  the  stimulus 
of  original  work  in  Geometry.  The  energy  expended  and 
enthusiasm  displayed  in  the  solution  of  exercises  is  in  sharp 
contrast  with  their  apathy  toward  the  study  of  theorems  in 
the  text. 

That  the  student  may  arrive  by  the  shortest  path  to  the 
point  where  the  real  development  of  his  mental  power  begins, 
where  his  interest  in  the  independent  solution  of  original 
exercises  becomes  an  active  part  of  his  school  life,  it  is  of  the 
first  importance  that  the  theorems  of  the  text  be  so  classified 
and  demonstrated  as  to  offer  his  attempts  to  master  them  the 
least  resistance.  It  is  of  equal  importance  that  these  exer- 
cises should  be  so  carefully  graded  as  to  stimulate  and  not 
discourage  the  pupil,  and  that  some  method  of  systematically 
attacking  and  solving  •  problems  should  be  devised  and  pre- 
sented for  his  assistance.  Nothing  is  presented  in  the  pages 
that  follow  that  has  not  stood  the  actual  test  of  class-room 
experience  for  many  years. 

In  accordance  with  these  essentials  of  a  text  for  use  in 
Geometry  classes,  we  call  special  attention  to  the  following 
features  of  this  work: 

First.     The  classification  of  Definitions  and  Axioms. 

Second.  The  arrangement  into  groups  of  Theorems  relating 
to  the  same  topic.  For  example,  theorems  concerning  isosceles 
triangles  in  the  "Isosceles  Triangle  Group";  congruent  tri- 
angles in  the  "Congruent  Triangle  Group";  comparison  oi 
areas  in  the  "Areal  Katio  Group." 

v 


vi  PLAN  AND  SCOPE 

Third.  The  arrangement  of  original  exercises,  second  in 
importance  only  to  the  grouping  of  theorems.  The  exercises 
are  not  only  attached  to  the  groups  of  theorems  upon  which 
their  solution  depends,  but  are  graded  according  to  their 
degree  of  difficulty. 

Fourth.  The  elimination  of  all  theorems  not  essential  to  a 
clear  understanding  of  the  principles  of  Geometry;  hence, 
a  number  of  theorems  found  with  their  proofs  in  the  usual 
text  are  given  as  exercises. 

Fifth.  The  compactness  of  each  group  and  the  simplicity 
and  clearness  of  the  demonstrations.  With  few  exceptions, 
not  more  than  three  or  four  recitations  are  needed  to  master 
any  one  of  the  groups. 

Sixth.  The  helpful  suggestions  as  to  the  method  of  solving 
original  exercises. 

Seventh.  A  simple  statement,  with  illustrations,  of  the  close 
connection  between  Algebra  and  Geometry;  classification  of 
principles  of  the  analogy  between  geometric  and  algebraic 
work  into  Indeterminate,  Determinate,  and  Overde terminate 
groups. 

In  the  grouping  of  Theorems  and  in  the  arrangement  and 
grading  of  Problems,  there  is  constantly  employed,  as  a 
valuable  aid  to  the  student,  the  principle  of  the  Association 
of  Similars. 

In  the  treatment  of  the  Spherical  Geometry,  much  space 
and  time  are  saved  by  utilizing  the  common  properties  of  the 
plane  and  sphere-surface ;  and  thus  transferring,  where  pos- 
sible, the  theorems  and  proofs  of  the  plane  to  the  sphere- 
surface. 

It  is  believed  that  this  plan  not  only  gives  the  pupil  a 
clearer  comprehension  of  the  unity  of  the  subject  than  he  will 
otherwise  obtain,  but  that  it  is  also  eminently  suggestive  of 
the  unlimited  possibilities  of  the  extension  of  geometric  truths 
to  other  surfaces. 


SUGGESTIONS  TO  TEACHERS 

Just  as  each  brick  in  a  building  rests  upon  a  brick  below  it, 
the  whole  superstructure  standing  upon  a  securely  laid  founda- 
tion, so  the  proof  of  each  Theorem  in  Geometry  rests  upon  the 
proof  of  the  preceding  Theorem,  which  in  turn  must  finally 
depend  upon  the  Definitions  and  Axioms.  Definitions  and 
Axioms,  therefore,  must  be  carefully  studied  and  thoroughly 
understood. 

For  example,  to  prove  that  the  J3i sector  of  the  Vertex  Angle 
of  an  Isosceles  Triangle  is  identical  with  the  Altitude  (Group 
IV,  1  6),  we  must  know  the  definitions  of  the  following  words : 
Bisector,  Vertex  Angle,  Isosceles  Triangle,  and  Altitude.  To 
understand  the  meaning  of  Altitude,  we  must  know  what  a 
Perpendicular  is,  wdiich,  in  turn,  requires  that  we  know  the 
meaning  of  a  Bight  Angle. 

The  student,  therefore,  before  attempting  the  proof  of  a 
Theorem,  should  be  made  to  understand  the  meaning  of  each 
and  every  term  found  in  the  Hypothesis  and  Conclusion  of 
the  Theorem. 

To  cultivate  the  habit  of  defining  terms  before  using  them, 
it  affords  the  student  valuable  if  not  indispensable  exercise  to 
require  of  him  frequent  definitions  of  all  the  terms  found  in 
the  Theorems  and  Corollaries,  particularly  of  the  first  six 
groups. 

Beginners  have  great  difficulty  in  keeping  in  mind  the  parts 
given  in  the  Hypothesis  distinct  from  those  given  in  the 
Conclusion.  During  the  process  of  demonstration  they  con- 
fuse what  was  to  be  proved  with  what  was  given.      It  will 

vii 


Tlii  SUGGESTIONS  TO  TEACHERS 

relieve  this  confusion  if  they  form  the  habit  of  marking  the 
parts  given  in  the  Hypothesis  by  the  usual  symbols,  i.e. : 

First.  If  lines  are  given  parallel,  by  drawing  the  symbol  Q 
across  the  lines. 

Second.  If  one  line  is  greater  than  another,  by  drawing  the 
symbol  >. 

Third.  If  angles  are  given  equal,  by  making  the  proper 
symbol  at  the  vertex,  just  within  the  sides  of  the  angle. 

Fourth.  All  parts  mentioned  in  the  Conclusion  may  be 
marked  with  an  X  • 

In  drawing  triangles,  unless  the  triangle  is  Isosceles,  Equi- 
lateral, or  Right,  it  is  well  to  adopt  the  following  rule : 

First.     Draw  the  base  line. 

Second.    Find  approximately  its  mid-point. 

Third.  Place  the  pencil  or  chalk  to  the  left  a  convenient 
distance  and  erect  an  imaginary  perpendicular. 

Fourth.  At  any  suitable  point  on  this  perpendicular  select 
the  vertex  of  the  vertex  angle,  from  which  draw  the  sides  of 
the  triangle. 

In  this  way  the  beginner  may  avoid  the  pitfalls  of  giving 
special  proofs  for  the  Isosceles,  Equilateral,  and  Right  Triangle 
that  will  not  apply  to  the  general  Triangle. 

In  drawing  Parallelograms,  unless  a  Rectangle,  Square,  or 
Rhombus  be  given  in  the  Hypothesis,  always  construct  a 
Rhomboid. 

At  the  beginning  of  the  course,  by  written  exercises  and 
much  blackboard  work,  familiarize  the  student  with  the  use 
of  symbols  and  abbreviations  ;  also  with  the  freehand  drawing 
of  the  altitudes  of  obtuse  triangles. 

After  the  tirst  month's  work  require  frequent  written  exami- 
nations, insisting,  as  the  course  advances,  that  close  attention 
be  paid  to  the  form  in  which  written  work  is  presented. 
Before  the  close  of  the  course  in  Plane  Geometry  the  student 
should  be  able  to  present  his  examination  papers  in  the  com- 
pact form  found  in  the  text. 


CONTENTS 

PAes 

Symbols : zi 

Abbreviations xii 

Definitions 1-9 

Extension 1 

Figures 2 

Angles 4 

The  Circle  and  the  Locus    .      ' 8 

Axioms 9 

Three  Preliminary  Theorems  on  Inequality     ....  10 

General  Terms 11 

Ten  Easy  Exercises  in  Geometrical  Dramming   ....  13 


PLANE   GEOMETRY 

I.     The  Group  on  Adjacent  and  Vertical  Angles  .         .  19 

11.     The  Parallel  Group .23 

III.  The  (2w-4)  Right  Angles  Group        ....  30 

IV.  The  Group  on  Isosceles  and  Scalene  Triangles        .  35 
V.     The  Group  on  Congruent  Triangles    ....  46 

VI.     The  Group  on  Parallelograms 53 

VII.     The  Group  on  Sum  of  Lines  and  Mid-joins          .         .  61 

VIII.     The  Group  on  Points  —  Equidistant  and  Random        .  69 

Nine  Illustrations  of  Elementary  Principles  of  Loci      .  74 

IX.     The  Group  on  the  Circle  and  its  Related  Lines     .  78 

X.     The  Group  on  Concurrent  Lines  of  a  Triangle         .  91 

Summary  of  Triangular  Relations         ....  99 

XI.     The  Group  on  Measurement 100 

Ratio  and  Proportion    .......  101 

Method  of  Limits 101 

ix 


X  CONTENTS 

PAOK 

XII.     The  Group  on  Measurement  of  Anoles       .         .         .  Ill 

Hints  to  the  Solution  of  Original  Exercises  .        .         .  120 

Illustration  of  the  Method  of  solving  Original  Problems  122 

Theorems  of  Special  Interest 127 

Classification  of  Problems  —  Indeterminate,  Determi- 
nate, and  Overdeterminate 130 

XIII.  The    Group    on    Areas    of    Rectangles    and    Other 

Polygons 138 

XIV.  The  Pythagorean  Group 149 

XV.     The  Group  on  Similar  Figures 160 

XVI.     The  Group  on  Areal  Ratios 176 

XVII.     The  Group  on  Linear  AppLiCATiiON  of  Proportion     .  183 
XVIII.     The  Group  on  Circumscribed  and  Inscribed  Regular 

Polygons 210 

XIX.     The  Group  on  the  Area  of  the  Circle       .         .         .  220 

XX.     The  Group  on  Concurrent  Transversals  and  Normals  228 


SOLID   GEOMETRY 

XXL     The  Group  on  the  Plane  and  its  Related  Lines       .  233 

XXII.     The  Group  on  Planal  Angles 253 

XXIII.  The  Group  on  the  Prism  and  the  Cylinder        .         .  266 

XXIV.  The  Group  on  the  Pyramid  and  the  Cone  .        .        .  287 
XXV.     The  Group  on  the  Sphere 311 

XXVI.     The    Group   on    Geometry   of   the    Sphere    Surface; 

Briefly',  Spherical  Geometry           ....  328 
Correspondence  between  Plane  and  Spherical  Geometry  331 
Summary  of  Propositions  Common  to  Plane  and  Spheri- 
cal Geometry 332 

Notes  and  Biographical  Sketches 347 

Index .         .  349 


SYMBOLS 


1.  Letter  points  with  capitals. 

2.  Letter  lines  with  small  letters. 

3.  Name  angles,    when  there   is 
italicized. 


no   ambiguity,   with   small   letters 


f^ 

(Greek  word  kentron)  center  in  general. 

Ki 

Center  of  inscribed  circle,  i.e.  in-center. 

Ke 

Center  of  escribed  circle,  i.e.  ex-center. 

4.    Points 

Ka 

Center  of  circumscribed  circle,  i.e.  circum-center. 

Ko 

Ortho-center  (intersection  of  altitudes). 

^9 

Centroid  or  Center  of  Gravity  of  triangle  (inter- 
section of  medians). 

■  Sides  of  a  triangle  :  Use  small  letters  corresponding  to 

the  capitals  at  the  vertices  opposite  the  respective 

sides  ;  a  opposite  A,  etc. 

_L 

Perpendicular,  or  "  is  perpendicular  to." 

5.   Lines 

Mid  ±  Mid-perpendicular. 

/ 

Oblique. 

II 

Parallel. 

O 

Circumference  or  circle. 

r\ 

Arc. 

6.   Angle 

Rt.  Z 

Angle  in  general. 
Right  angle. 

7.   Triangle  • 

A 
Rt.  A 

Triangle  in  general. 
Right  triangle. 

4-side 

o 
1    1 

o 

Quadrilateral  in  general. 

8.    Quadri- 
lateral 

Parallelogram  or  rhomboid. 

Rectangle. 

Rhombus. 

ID 

Square. 

■••• 

Therefore. 

Since  or  because. 

,^ 

Similar. 

9.    Miscel- 
laneous 

^ 

Congruent. 
Identical. 

> 

Is  greater  than. 

< 

Is  less  than. 

=: 

Approaches  as  a  limit. 

For  the  plura 

il  add  the  letter  s. 

The  cancella 

tion  across  a  symbol  means  "not,"  e.g. :  >  means  not 

greater  than ;  i 

-  means  not  parallel. 

ABBREVIATIONS 


Alt. 

Alternate. 

0pp. 

Opposite. 

Ax. 

Axiom. 

Prop. 

Proposition. 

Adj. 

Adjacent. 

Prob. 

Problem. 

Cone. 

Conclusion. 

Q.E.D. 

Quod  erat  demonstrandum 

Const. 

Construction. 

(which  was  to  be  proved). 

Cor. 

Corollary. 

Q.E.F. 

Quod  erat  faciendum 

Corr. 

Corresponding. 

(which  was  to  be  done). 

Def. 

Definition, 

Supp. 

Supplemental. 

Dem. 

Demonstration. 

Th. 

Theorem. 

Ex. 

Exercise. 

Vert. 

Vertical. 

Ext. 

Exterior. 

V. 

Vide  (see). 

Hyp. 

Hypothesis. 

q.v. 

Quod  vide  (which  see). 

Horn. 

Homologous. 

cf. 

Compare. 

Int. 

Interior. 

r. 

Radius. 

n-gon. 

Polygon. 

Groups,  Theorems,  and  Corollaries  are  read  as  follows :  II.  1.  a  means 
Group  II,  Theorem  1,  Corollary  a. 


zii 


THE  ELEMENTS  OF  GEOMETRY 


DEFINITIONS 

EXTENSION 

The  Definition  of  a  mathematical  term  is  its  explanation  in 
words  familiar  to  the  student. 

The  test  of  a  complete  definition  is  that  the  subject  and 
predicate  may  be  interchanged  without  affecting  the  truth  of 
the  statement. 

Space  extends  about  us  on  every  side.  Every  material  object 
occupies  a  portion  of  this  space.  This  portion  of  space  is  called 
a  geometrical  solid  or  simply  a  Solid.  The  only  properties  of 
the  solid  with  which  geometry  is  concerned  are  its  form  and 
size,  and  its  position  with  reference  to  other  solids. 

A  boundary  of  a  solid  is  called  its  Surface.  It  is  no  part  of 
the  solid,  and  therefore  has  but  two  dimensions :  length  and 
breadth. 

A  boundary  of  a  surface  is  called  a  Line.  It  is  no  part  of 
the  surface,  and  has  therefore  but  one  dimension :  length. 

A  boundary  of  a  line  is  called  a  Point.  A  point  has  no 
dimension,  but  position  only. 

A  point,  line,  or  surface  that  divides  any  magnitude  into 
two  equal  parts  is  called  the  Bisector  of  that  magnitude. 

Any  definite  portion  of  a  line  is  called  a  Line-segment. 

A  Straight  Line  is  a  line  that  lies  evenly  between  its  extreme 
points;  that  is,  if  the  ends  of  one  segment  may  be  placed 
upon  the  ends  of  a  second  segment,  the  segments  must  coincide 
throughout  their  whole  extent. 

1 


2  THE  ELEMENTS  OF  GEOMETRY 

A  straight  line  conuecting  any  two  points  is  called  the  Join 
of  thos^  pojots. 

A  Broken  Line  is  a  series  of  joins,  any  two  consecutive  joini 
having  one  point  in  common. 

A  Curved  Line  is  a  line  such  that  no  segment  of  it  is  straight. 

Concurrent  Lines  are  lines  passing  through  the  same  point. 

A  Transversal  is  any  line  intersecting  a  number  of  other 
lines. 

Note.  —  in  the  text  the  word  line  is  used  to  mean  a  straight  line. 

FIGURES 

A  Figure  or  Complex  is  any  collection  of  points,  lines,  or 
points  and  lines. 

Similar  figures  are  figures  having  the  same  shape. 

Equal  figures  are  figures  having  the  same  size. 

Congruent  figures  are  figures  having  the  same  shape  and  the 
same  size. 

The  Test  of  Congruency  is  that  one  figure  may  be  plaqed  on 
the  other  so  that  every  part  of  the  first  will  coincide  with  the 
corresponding  part  of  the  second. 

Two  figures  thus  placed  are  said  to  be  in  Coincident  Super- 
position. 

A  Plane  is  a  surface  such  that  if  any  two  of  its  points  be 
joined  by  a  straight  line,  this  line  must  lie  wholly  within  the 
surface. 

A  Plane  Figure  is  one  all  the  parts  of  which  lie  in  the  same 
plane. 

Note.  — All  figures  hereafter  defined  are  assumed  to  be  plane  figures. 

A  Polygon  is  a  portion  of  a  plane  bounded  by  a  closed  broken 
line  called  its  Perimeter. 

When  only  the  form  is  considered,  the  word  polygon  is  fre- 
quently used  to  mean  the  perimeter  of  the  polygon. 


DEFINITIONS  3 

A  three-sided  polygon  is  called  a  Triangle. 

A  four-sided  polygon  is  called  a  Quadrilateral  or  4-8ide. 

A  five-sided  polygon  is  called  a  Pentagon. 

A  six-sided  polygon  is  called  a  Hexagon. 

A  seven-sided  polygon  is  called  a  Heptagon. 

An  eight-sided  polygon  is  called  an  Octagon. 

A  nine-sided  polygon  is  called  a  Nonagon. 

A  ten-sided  polygon  is  called  a  Decagon. 

A  twelve-sided  polygon  is  called  a  Dodecagon. 

A  fifteen-sided  polygon  is  called  a  Pentadecagon. 

The  Vertices  of  a  polygon  are  the  points  in  which  its  consecu- 
tive sides  meet. 

A  Diagonal  of  a  polygon  is  the  join  of  any  two  non-consecu- 
tive vertices. 

A  Regular  Polygon  is  a  polygon  whose  angles  are  equal  and 
whose  sides  are  equal. 


Ex.  1.   What  is  the  test  of  a  complete  definition  ? 

Ex.  2.   Apply  this  test  to  the  definition  of  a  line-segment. 

Ex.  3.  Define  (a)  Bisector.  (&)  Line,  (c)  JoiU'  (d^  Concurrent 
lines,     (e)  Transversal. 

Ex.  4.    Define  a  straight  line. 

Can  you  place  two  equal  portions  of  a  barrel  hoop  in  such  a  way  that 
the  ends  of  one  will  coincide  with  the  ends  of  the  other,  but  the  portions 
(or  segments)  themselves  will  not  coincide  ? 

Can  you  place  them  so  that  they  will  coincide  ? 

Since,  then,  equal  segments  of  a  curve  may  or  may  not  be  made  to 
coincide,  what  is  the  word  to  emphasize  in  the  definition  of  a  straight 
line  ? 

Ex.  5.   Draw  three  concurrent  lines. 

Ex.  6.    Draw  three  non-concurrent  lines. 

Ex.  7.   Draw  a  transversal  to  two  lines. 

^x.  8.   What  are  congruent  figures  ? 

Ex,  9.    What  is  the  test  of  congruency  ? 

Ex.  10.    Give  illustrations  of  equal,  similar,  and  congruent  figured. 

Ex.  11.   What  are  the  two  dimensions  of  a  surface  ? 

Ex.  12.   Define  a  plane. 

Ex.  13.  Using  your  ruler  as  a  straightedge,  show  that  the  top  of  your 
desk  is  a  plane. 


4  THE  ELEMENTS  OF  GEOMETRY 

ANQLBB 

An  Angle  is  a  figure  formed  bj  two  lines  that  meet ;  the  lints 
being  called  the  Sides  of  the  angle. 

The  point  of  meeting  of  the  sides  of  the  angle  is  called  the 
Vertex. 

The  usual  Method  of  Reading  an  angle  is  to  read  the  three 
letters  on  the  sides,  placing  the  letter  at  the  vertex  between 
the  other  two.  If  there  is  no  ambiguity,  the  angle  may  be  read 
by  the  letter  at  the  vertex. 

If,  when  the  sides  of  an  angle  are  produced,  all  the  angles 
formed  are  equal,  each  angle  is  called  a  Right  Angle,  and  the 
lines  are  said  to  be  Perpendicular  to  each  other. 

If  a  line  bisects  a  second  line  and  is  also  perpendicular  to  it, 
the  first  line  is  called  the  Mid-normal  q 

or  Mid-perpendicular  to  the  second. 

ABC  is  an  angle.     Its  vertex  is  B.    If 
the  side  AB  is  produced  to  E  and  the  side 

CB  is  produced  to  F,  and  if  angles  ABC^    ^ \.~ e 

CBE,  EBF,  and  ABF  are  all  equal,  then 
they  are  right  angles,  and  the  line  CF  is 
perpendicular  to  the  line  AE.  If  the  line 
CF  also  bisects  the  line  AE,  it  is  the  Mid- 
normal  to  AE.  j* 

By  the  distance  of  a  point  from  a  line  is  meant  the  perpen- 
dicular distance ;  by  the  distance  between  two  lines,  the  per- 
pendicular distance. 

Classes   of  Angles 
(a)  As  to  their  Algebraic  Sign 

An  angle  may  be  considered  as  generated  by  the  revolution 
of  one  line  about  a  fixed  point  in  a  second  line. 

The  rotating  line,  when  it  comes  to  rest,  is  called  the  Termi- 
nal Line. 

The  fixed  line  is  called  the  Initial  Line. 


DEFINITIONS 


If  the  rotating  line  moves  anti-clock- 
wise, the  angle  generated  we  call  Positive ; 
if  clockwise,  Negative. 

If  the  rotating  line,  moving  from  a 
position  coincident  with  the  initial  line, 
complete  a  revolution,  it  generates  four 
right  angles,  or  a  Perigon ;  if  half  a  revo- 
lution, it  generates  two  right  angles,  or  a 
Straight  Angle. 

Note.  —The  size  of  an  angle  does  not  depend 
upon  the  length  of  its  sides. 

Note.  — In  finding  the  sum,  Z.a  -f-  Z&,  of  2/:!, 
place  the  initial  line  and  vertex  of  Z&  on  the 
terminal  line  and  vertex  of  Aa. 

Then  generate  the  Z  &  by  the  rotation  indicated 
by  its  sign. 

Then  the  angle  between  the  initial  line  of  Za 
and  the  terminal  line  of  Z  &  is  Z  a  +  Z  6. 

The  Relative  Direction  of  one  line  with 
respect  to  another  is  the  angle  that  the 
first  line  makes  with  the  second,  both 
the  size  and  sign  of  the  angle  being  con- 
sidered. 

The  direction  of  rotation  of  the  line 
generating  the  angle  will  be  considered 
positive  unless  the  contrary  is  stated. 

Relative  direction  will  be  the  only  direc- 
tion considered  in  this  book;   absolute  direction  will  not  be 
discussed. 

(h)  As  to  Size 

An  Acute  angle  is  an  angle  less  than  a  right  angle. 
An  Obtuse  angle  is  an  angle  greater  than  a  right  angle. 
An  obtuse  angle  equal  to  two  right  angles  is  called  a  Straight 
angle. 

An  Oblique  angle  is  any  angle  that  is  not  a  right  angle. 


6  THE  ELEMENTS  OF  GEOMETRY 

Wheu  two  angles  are  both  acute  or  both  obtuse,  they  are 
said  to  be  of  the  Same  Kind. 

If  the  sum  of  two  angles  equals  one  right  angle,  they  are  called 
Complemental  angles. 

If  the  sum  of  two  angles  equals  two  right  angles,  they  are 
called  Supplemental  angles. 

,  (c)  As  to  Location  * 

Adjacent  angles  are  angles  that  have  a  common  side  and  a 
common  vertex. 

Vertical  angles  are  the  alternate  angles  formed  by  two  lines 
that  cross  each  other. 

Angles  formed  by  a  Transversal  with  Two  Other  Lines 

Angles  within  the  two  lines  crossed  by  a  transversal  are 
called  Interior  angles ;  angles  without,  Exterior  angles. 

Non-adjacent  angles  on  the  same  side  of  the  transversal  are 
called  Corresponding  angles. 

There  are  three  classes  of  corresponding  angles :  Corresponding 
Interior,  Corresponding  Exterior,  and  Corresponding  Exterior- 
interior  angles. 

Corresponding  exterior-interior  angles  are  Corresponding 
Non-adjacent  angles,  one  of  which  is  interior  and  the  other 
exterior. 

Alternate  angles  are  angles  on  opposite  sides  of  the  trans- 
versal that  do  not  have  the  same  vertex. 

There  are  three  classes  of  alternate  angles :  Alternate  Interior, 
Alternate  Exterior,  and  Alternate  Exterior-interior  angles. 

Alternate  exterior-interior  angles  are  alternate  angles,  one 
exterior  and  the  other  interior,  but  not 
having  a  common  vertex. 

Point  out  in  the  adjoining  figure  the  following 
angles  :  acute,  obtuse,  oblique,  supplemental,  ad- 
jacent, opposite,  of  the  same  kind,  corr.  ext.,  corr. 
int.,  corr.  ext.-int.,  alt.  ext.,  alt.  int.,  alt.  ext. -int. 


DEFINITIONS  7 

A  Postulate  is  a  construction  admitted  to  be  possible. 

Post.  There  may  always  be  drawn  a  pair  of  lines  that  make 
equal  corresponding  exterior-interior  angles  with  any  trans- 
versal whatever  cutting  them. 

Two  lines  are  Parallel  when,  if  cut  by  any  transversal  what- 
ever, the  corresponding  exterior-interior  angles  are  equal. 

Ex.  14.  The  ruler  can  be  so  placed  as  to  lie  wholly  on  the  stovepipe. 
Why,  then,  is  not  the  stovepipe  a  plane  ? 

Ex.  15.  Show  that  any  two  parallels  have  the  same  direction  with  re- 
spect to  any  transversal. 

Ex.  16.  Define:  (a)  Angle.  (6)  Right  angle,  (c)  Perpendiculars. 
(d)  Perpendicular  bisector  (or  mid-normal).  (e)  Oblique  angles. 
(/)  Angles  of  the  same  kind. 

Ex.  17.    What  is  the  complement  of  two  fifths  of  a  right  angle  ? 

Ex.  18.    What  is  the  supplement  of  two  fifths  of  a  right  angle  ? 

Ex.  19.   Name  the  angles  formed  by  the  clock  hands  : 

(a)   10  minutes  after  three  o'clock,      (c)    20  minutes  after  three  o'clock. 
(6)    15  minutes  after  three  o'clock,      (d)   30  minutes  after  three  o'clock. 

Ex.  20.  Draw  two  oblique  angles ;  draw  two  angles  of  the  same 
kind. 

Ex.  21.  Draw  two  angles  that  are  :  (a)  Adjacent.  (6)  Adjacent  and 
complemental.  (c)  Adjacent  and  supplemental,  (d)  Non-adjacent  and 
supplemental. 

Ex.  22.  An  angle  is  one  fifth  its  complement.  What  is  the  value  of 
the  angle  ? 

Ex.  23.  Two  angles  are  equal  and  at  the  same  time  complemental. 
What  is  their  value  ? 

Ex.  24.  Two  angles  are  equal  and  at  the  same  time  supplemental. 
What  is  their  value  ? 

Ex.  25.  Draw  two  angles  that  have  a  common  side  but  not  a  common 
vertex. 

Ex.  26.  Draw  two  angles  that  have  a  common  vertex  but  not  a  common 
side. 

Ex.  27.    Draw  two  lines  so  that  a  transversal  crossing  them  makes : 

(a)    Corresponding  exterior-interior  angles  equal. 

(&)   Corresponding  interior  angles  equal,     (c)    All  the  angles  equal. 

Ex.  28.  In  case  (a)  of  the  preceding  question  what  other  angles  are 
equal  ? 


8  THE  ELEMENTS  OF   GEOMETRV 

THE  CIRCLE  AND   THE  LOCUS 

A  Circle  is  a  portion  of  a  plane  bounded  by  a  curved  line, 
called  the  Circumference,  every  point  of  which  is  equidistant 
from  a  point  called  the  Center. 

Any  line  from  the  center  to  the  circumference  is  called  a 
Radius. 

Ciiclefi  having  the  same  center  are  Concentric  Circles. 

A  Locus  is  a  line  or  a  complex,  all  points  of  which  possess  a 
common  property  that  does  not  belong  to  any  point  without 
this  line  or  complex.  E.g.  it  is  the  common  property  of  every 
point  in  the  circumference  of  a  circle  that  is  a  radial  dis- 
tance from  the  center;  the  circumference  is  the  locus  of  all 
points  that  are  a  radial  distance  from  the  center. 

Ex.  29.    Do  the  two  lines  meet  ? 

Ex.  30.    In  case  (6)  what  other  angles  are  equal  ? 

Ex.  31.    Do  the  two  lines  meet  ? 

Ex.  32.  In  case  (c)  do  the  two  lines  meet  ?  The  angles  in  case  (c)  are 
all  of  what  kind  ? 

Ex.  33.    How  may  an  angle  be  considered  to  be  generated  ? 

Ex.  34.    Define  a  negative  angle. 

Ex.  35.  The  hands  of  the  clock  are  together  at  twelve  o'clock.  If  the 
hour  hand  is  stationary,  what  is  the  algebraic  sign  of  the  angle  formed  by 
the  hands  at  ten  minutes  after  twelve  ? 

Ex.  36.  If  the  hour  hand  is  stationary  and  the  minute  hand  is  moved 
to  the  left,  what  is  the  algebraic  sign  of  the  angle  formed  at  ten  minutes 
before  twelve  ? 

Ex.  37.  What  is  the  sign  of  the  angle  generated  by  a  line  from  the 
observer  to  the  moon  from  moonrise  to  moonset  ? 

Ex.  38.  To  mariners  what  is  the  fixed  or  known  line  ?  What  instru- 
ment on  every  sliip  serves  to  determine  this  line  ? 

Ex.  39.  If  a  ship  is  sailing  southeast,  what  angle  does  its  course  make 
with  this  known  line  ? 

Ex.  40.  What,  then,  is  the  relative  direction  of  the  ship  with  respect  to 
this  fixed  line  ? 

Note.  —  Answer  all  questions  concerning  angles  in  terms  of  a  right 
angle  until  the  word  degree  has  been  defined. 

Ex.  41.    What  angle  is  one  fifth  its  supplement  ? 


DEFINITIONS  9 

The   truths  of   geometry  are  expressed  by  its   definitions, 
axioms,  and  theorems. 

An  Axiom  is  a  statement,  the  truth  of  which  is  assumed. 

AXIOMS 

1.  Things  equal  to  the  same  thing  or  equal  things  are  equal 

to  each  other. 

2.  If  equals  be  added  to  or  subtracted  from  equals,  the  results 

will  be  equal. 

3.  If  equals  be  multiplied  or  divided  by  equals,  the  results 

will  be  equal. 

4.  The  whole  equals  the  sum  of  its  parts. 

Direct  Inference :  The  whole  is  greater  than  any  of  its  parts. 

5.  The  intersection  of  two  lines,  straight  or  curved,  fixes  the 

position  of  a  point. 
E.g.  the  intersection  of  the  latitude  and  longitude  of  a  ship 
at  sea  determines  its  position. 

6.  Two  points  fix  the  position  of  a  line. 

E.g.  if  a  railroad  extending  in  a  straight  line  passes  two 
stations  whose  positions  are  known,  the  railroad  is  also 
determined  or  fixed  in  position. 

7.  A  point  and  the  direction  of  -a  straight  line  determine  its 

position. 
E.g.  if  a  railroad  extends  northwest  and  passes  a  known 
station,  the  position  of  the  railroad  is  known. 

a.  From  a  point  without  a  line  but  one  perpendicular 

can  be  drawn  to  the  line. 

b.  At  a  point  in  a  line  but  one  perpendicular  can  be 

drawn  to  the  line. 

8.  Through  a  point  one  line  and  only  one  can  be  drawn 

parallel  to  a  given  line. 

9.  Any  figure  may  be  transferred  from  one  position  in  space 

to  any  other  without  change  of  size  or  shape. 
10.   If  there  be  but  one  x  and  one  y,  then,  from  the  fact  that 
X  is  y,  it  necessarily  follows  that  y  is  x. 


lU  THE  ELEMENTS  OF   GEOMETRY 

THREE  PRELIMINARY   THEOREMS   ON   INEQUALITY 

These  propositions  are  given  in  many  texts  as  axioms. 
They  are  proved,  however,  in  the  leading  treatises  on  algebra. 

1.  If  unequals  are  added  to  unequals  in  the  same 
sense,  the  results  will  he  unequal  in  the  same  sense, 

Dem. :  Suppose  a > 6,  and  c>  e. 

To  prove  a-^c>h-\-e. 

a'>h.    ,',  a  =  h-\-  some  quantity,  say  x. 
c> c.     .\  c  —  e-\- some  quantity,  say  y. 
Thus  a=-h  +  Xy 

c  =  e  +  y. 
.'.  a  +  c  =  5  +  e  +  a;  +  2^.  (Ax.  2.) 

That  is,  a  -f  c  >  &  +  e. 

Q.E.D. 

2.  If  equals  are  added  to  unequals,  the  results  are 
unequal  in  the  same  sense, 

Dem. :  Suppose  a  >  h,  and  c  =  e 

To  prove  a  -f  c  >  6  +  e. 

a  >  6.     .*.  a  =  &  +  some  quantity,  say  x. 

Then  a  =  &  -}- », 

and  c  =  e. 

.\  a  +  c  =  b  •^e-\'X.  (Ax.  2.) 

.*.  a-f  c>  &4-e. 

Q.E.D. 

3.  If  equals  are  subtracted  from  unequals,  the  results 
are  unequal  in  the  same  sense. 

The  proof  of  this  proposition  is  precisely  similar  to  that 
of  Theorem.  2. 

Ex.  42.  What  is  the  supplement  of  the  angle  between  the  hands  of  a 
clock  at  five  o'clock  ? 

Ex.  43.  What  is  the  complement  of  'the  angle  in  the  preceding  ques- 
tion ?    (v.  Negative  angles.) 


DEFINITIONS  11 

GENERAL    TERMS 

A  Theorem  is  a  statement  to  be  proved.  It  consists  of  two 
parts : 

The  Hypothesis  (Hyp.),  or  supposition  or  premise. 
The  Conclusion  (Cone),  or  what  is  asserted  to  follow  from 
the  hypothesis. 

A  Proof  is  a  course  of  reasoning  by  which  the  truth  of  a 
theorem  is  established. 

The  Converse  of  a  theorem  is  obtained  by  interchanging  the 
hypothesis  and  conclusion  of  the  original  theorem. 
Theorem :  If  ^  is  B,  then  G  is  E. 
Converse :  If  (7  is  E,  then  A  is  B. 
Note.  — The  converse  is  sometimes  called  the  indirect  theorem. 
The  Contradictory  of  a  theorem  is  true  if  the  theorem  is  false, 
and  vice  versa. 

Theorem  :  If  ^  is  B,  then  C  is  E. 
Contradictory :  If  A  is  B,  then  C  is  not  E. 

The  Opposite  of  a  theorem  is  obtained  by  making  both  the 
hypothesis  and  conclusion  negative. 

Theorem :  If  A  is  B,  then  C  is  E. 
Opposite :  If  ^  is  not  B,  then  C  is  not  E. 

A  Reciprocal  theorem  is  formed  by  replacing,  when  possible, 
in  the  original  theorem,  the  words  "  point  by  line,"  "  line  by 
point,"  "  angles  of  a  triangle  by  the  opposite  sides  of  the  tri- 
angle," "  sides  of  a  triangle  by  the  opposite  angles  of  the  tri- 
angle," "opposite  angles  of  a  4-side"  by  "the  opposite  sides  of 
a  4-side,"  etc. 

Note.  —  For  every  statement  in  a  proof  a  reason  must  be  given. 
This  reason  must  be : 

1.  A  hypothesis.  3.   A  definition. 

2.  A  construction.  4.   An  axiom. 
5.   A  previously  established  theorem  or  corollary. 

A  Corollary  is  a  subordinate  statement  deduced  from,  or 
.suggested  by,  the  main  statement  or  its  proof. 


12  THE  ELFMENTS  OF  GEOMETRY 

A  iH-oblem  requires  the  construction  of  a  geometric  figure 
that  will  satisfy  given  conditions. 

The  Solution  of  a  Problem  consists  of  four  parts : 

First:  The  Analysis,  or  course  of  reasoning  by  which  the. 
method  of  constructing  the  required  figure  is  discovered  or 
rediscovered. 

Note.  — The  analysis  of  problems  is  explained  in  full  under  the  article 
•'  Helps  to  the  Solution  of  Original  Problems,"  Group  XII. 

Second:  The  Construction  of  a  figure  after  the  method  has 
been  discovered. 

Third:  The  Proof  that  the  figure  satisfies  the  given  condi- 
tions. 

Fourth:  The  Discussion  of  the  changes  in  the  number  of 
figures  that  will  satisfy  the  given  conditions,  made  by  a  change 
in  the  size  of  the  given  magnitudes,  in  their  relative  position, 
or  in  both. 

A  Proposition  is  a  general  terra  applying  to  theorems  or 
problems. 

A  Scholium  is  a  remark  upon  a  particular  feature  of  a  propo- 
sition. 

A  Lemma  is  a  theorem  introduced  merely  to  be  used  in  the 
proof  of  one  immediately  following. 

Plane  Geometry  is  that  branch  of  mathematics  in  which  are 
considered  the  properties  of  magnitudes  lying  on  the  same 
plane. 

Suggestions  for  Class  Work  at  Blackboard 

In  lettering  a  fifjure  avoid  the  use  of  letters  that  have  the  same  sound  ; 
as  B,  Z>,  P,  and  T;  JWand  N,  etc. 

Place  uip  of  pointer  within  the  figure  you  are  to  read.  To  move  it 
from  point  to  point  is  confusing.  If  reading  a  line-segment,  place  tip 
about  the  middle  of  the  segment ;  if  an  angle,  place  it  near  the  vertex ; 
if  a  polygon  of  any  kind,  within  the  polygon. 

In  drawing  triangles  make  them,  unless  otherwise  directed,  scalene, 
with  the  angle  at  A  greater  than  the  angle  at  .B,  and  C  the  vertex  angle. 


TEN   EASY   EXERCISES   IN   GEOMETRICAL  DRAWING       13 


I 
II 

J 

VI 

IX 

X. 

III 

VJI 

IV 

VIIl 

TEN   EASY   EXERCISES   IN   GEOMETRICAL   DRAWING 

These  problems  are  introduced  at  this  point  to  familiarize 
the  student  with  the  use  of  the  ruler  and  compasses  and  with 
geometrical  terms.  The  constructions  are  simple,  and  serve  to 
illustrate  some  of  the  practical  applications  of  geometry. 

The  student  will  observe  that  the  constructions  in  Probs.  II, 
V,  VI,  IX,  and  X  are  direct  applications  of  the  constructions 
in  Prob.  I,  and  that  the  others  are  also  intimately  connected 
with  it.  Thus  Prob.  I  may  be  considered  the  string  from 
which  the  other 
problems  are  sus- 
pended. This  de- 
pendence is  shown 
by  the  adjacent 
diagram. 

The  proofs  for  these  constructions  are  to  be  given  as  soon  as  the 
necessary  theorems  have  been  established.  Eeferences  to  the  fol- 
lowing problems  are  made  in  footnotes  and  exercises  attached 
to  these  theorems. 

Summary 

Problem  I.  (a)  Bisect  a  given  line-segment, 

(b)  Erect  a  mid  -L  to  it. 
Problem  II.         Bisect  a  given  arc. 
Problem  III.       Bisect  a  given  angle. 
Problem  IV.        Ti'isect  a  given  right  angle. 
Problem  V.         Erect  a  A.  to  a  given  line  at  a  given 

point  in  the  line. 
Problem  VI.       Draw  a  ±  to  a  given  line  from,  a  given 

point  luithoiit  the  line. 
Problem  VII.      Inscribe  a  O  in  a  given  triangle. 
Problem  VIII.    Escribe  a  O  to  a  given  triangle. 
Problem  IX.        Circumscribe  a  O  to  a  given  triangle. 
Problem  X.         Find  the  center  of  a  given  O. 


14  THE  ELEMENTS  OF  GEOMETRY 

Problem  I.    (a)  To  bisect  a  given  line-sec/vient.  (b)  To 
erect  a  mid  Lto  a  given  line-segment 

Given.    The  line-segment  AB.  j 

Required,     (a)   To  bisect  AB.     (b)    To  ! 

erect  a  mid  J.  to  AB.  I 

A d B 


Const.     With  ^  as  a  center  and  a  r  =  AB,  ! 

describe  arcs  above  and  belpw  AB.  \ 

With  5  as  a  center  and  the  same  r,  de-  ^1^ 

scribe  arcs  above  and  below  AB.  E 

Let  these  arcs  intersect  above  the  line  in  C;  below  in  E. 

Draw  the  join  CE. 

Let  it  intersect  AB  in  0. 

Then  (a)  0  bisects  AB. 

(b)  CE  is  the  mid  ±  to  AB. 


Q.E.F. 


M\ 


Problem  II.     To  bisect  a  given  arc. 

Given.     The  arc  AB,  / 

Required.     To  bisect  the   arc 
AB. 


Const.     Draw  the  chord  AB. 
Construct  the  mid  J_  to  this  chord  by  Prob.  T. 
Let  this  mid  _L  intersect  the  arc  AB  in  M. 
Then  M  bisects  the  arc  AB. 


Q.E.F. 


Ex.  44.   In  what  line  do  you  find  all  the  house*  that  are  one  mile  from 
the  county  courthouse  ?         •  • 


TEN    EASY  EXERCISES   IN   GEOMETRICAL  DRAWING       15 

Problem  III.     To  hisect  an  angle.  j,^ 

Given.     Z  MAL.  .^rtlTrl™. \ 

Required.     To  bisect  Z  MAL.  ^""""""^-^.i.,^^ 

Const.     With  ^  as  a  center  and  any  r,  describe  an  arc  BC. 
Bisect  the  arc  BC  by  Prob.  II. 

The  join  of  A  and  the  mid-point  of  arc  BC  bisects  /.MAL. 

Q.E.F 


Problem  IV.    To  trisect  a  right  angle. 

c 
Given.     The  rt.  Z  A. 

Required.     To  trisect  rt.  Z  A. 


^^     /^^ 


/  V 


H 


Const.     Lay  off  on  AH  any  line-segment  AB. 
With  ^  as  a  center  and  AB  as  a  r,  describe  an  arc. 
With  J5  as  a  center  and  the  same  r,  describe  an  arc. 
Let  these  two  arcs  intersect  at  E. 
Draw  EA  and  bisect  Z  BAE  by  Prob.  III. 

ZBAT=Z.  TAE  =  Z.EAC. 


Q.E.F. 


Problem  V.    To  erect  a  perpendicular  to  a  line  at  a 
point  in  the  line.  . 

Given.     The  line  ^5  and  P  in  ^5.  ^j 

Required.     To  erect  a  X  to  AB                  j 
atP.  A      ^G        P         \i 'b 

Const.    Lay  off  P(7=PQ. 

Construct  the  mid  X  PF  to  CQ  by  Prob.  I. 

PP  is  the  required  X. 

Q.E.F. 


10  THE  ELEMENTS  OF  GEOMETRY 

Problem  VI.  To  draw  a  perpendicular  to  a  line  from 
a  point  without  the  line. 

Given.     The  line  ^JB  and  ^ 

the  point  P  without  AB. 

Required.     To  draw  a  ± 
from  P  to  AB.  a~^  fi X> b 

Const.     With  P  as  a  center  and  any  r  >  the  distance  from 

P  to  AB,  describe  an  arc  intersecting  AB  in  C  and  Q. 

Construct  the  mid  _L  PH  to  the  line-segment  CQ  by  Prob.  I. 

PH  is  the  required  ±. 

Q.E.F. 

Problem  VII.     To  inscribe  *  a  circle  in  a  triangle. 

G 
Given.     The  A.4JBC. 

Required.     To  inscribe  a  O 
in  A  ABC. 

"^  A 

Const.     Bisect  Z.  A  and  Z  jB  by  Prob.  III. 

Let  the  bisectors  intersect  at  some  point  K. 

Construct  a  J.  from  ^  to  ^B  by  Prob.  VI. 

With  ^  as  a  center  and  this  ±  as  a  r,  describe  a  O. 

This  0  will  be  inscribed  in  A  ABC. 

Q.E.F. 

Ex.  45.  What  is  the  locus  of  a  point  one  mile  from  a  given  point  ? 

Ex.  46.  What  is  the  locus  of  a  point  that  is  h  distant  from  a  given 
point  F7 

Ex.  47.  What  is  a  proof  or  demonstration  ? 

Ex.  48.  Into  what  two  parts  may  a  theorem  be  separated  ?    A  problem  ? 

Ex.  49.  Name  the  five  classes  of  reasons,  one  of  which  must  be  given 
for  every  statement  that  is  made  in  the  proof. 

*  A  O  is  inscrioed  in  a  A  when  the  sides  of  the  A  are  tangent  to  the  O 
(i.e.  touch  the  O  in  but  one  point).    See  p.  78. 


TEN  EASY   EXERCISES  IN   GEOMETlilCAL  DRAWING       17 

Problem  VIII.     To  escribe  ^  a  circle  to  a  ti'iangle. 

\P 

Given.     The  A  ABQ. 

Required.     To  escribe   a  O  to   the 
b.ABC. 


Const.     Produce  the  sides  h  and  a. 

Bisect  Z  BAE  and  Z  FBA  by  Prob.  III. 

Let  the  bisectors  intersect  in  some  point  K. 

Construct  a  _L  from  if  to  a  (or  to  IS)  produced. 

With  ^  as  a  center  and  this  _L  as  a  ?',  describe  a  O. 

This  O  will  be  escribed  to  A  ABC. 

Q.E.F. 

Problem  IX.     To  pass  a  circle  through  three  points. 

Given.     The  three  points,  A,  B,  and  C,     /  \  \ 

not  coUinear.  /  -'^''V^ 

if:-""\  ! 


Required.     To  pass  a  O  through  A,  B,    \              ;           \  / 
and  a  a\ r 7B 


I ,,' 


Const.     Draw  the  joins  AB  and  JBC. 

Construct  the  mid  Js  to  AB  and  BC  by  Prob.  I. 

Let  these  mid  Js  intersect  at  a  point  K. 

With  KsiS  a  center  and  a  r  =  KA,  KB,  or  KC,  describe  a  O. 

This  O  passes  through  A,  B,  and  (7. 

Q.E.F. 

Note.  —  If  A,  B,  and  C  are  connected,  the  above  G  is  said  to  be  cir- 
cumscribed to  A  ABC. 


1  To  escribe  a  circle  is  to  draw  it  tangent  %o  one  side  of  a  triangle  and 
to  the  other  two  sides  produced. 


18  THE  ELEMEIJTS  OF  GEOMETRY 

Problem  X.    To  find  the  center  of  a  given  circle. 


Given.    The  O  K, 

Required.     To  find  its  center. 


Const.     Take  any  three  points  in  the  O,  as  A,  B,  and  C. 

Draw  the  joins  of  any  two  of  these,  as  AB  and  BC. 

Construct  the  mid  Js  to  these  joins  by  Prob.  I. 

Let  these  mid  J§  intersect  at  some  point  K. 

K  is  the  center  required. 

Q.E.F. 

Note.  —  Prob.  X,  after  the  three  points  have  been  selected,  is  evidently 
identical  with  Prob.  IX. 

Note.  —  The  proofs  for  the  solution  of  these  problems  will  be  found  as 
follows :  — 

I.,V.,  VI.,p.  72. 

IL,  p.  79. 

III.,  p.  112. 

IV.,  p.  49,  with  Def.  p.  30. 

VII.,  p.  92,  X.,  1,  a. 

VIII.,  p.  93,  Scb. 

IX.  and  X.,  p.  94. 

Ex.  50.  What  is  the  opposite  of  a  theorem  ? 

Ex.  51.  State  the  opposite  of  Group  I,  Theorem  2. 


PLANE  GEOMETRY 

I.  THE  GROUP  ON  ADJACENT  AND  VERTICAL 
ANGLES 

PROPOSITIONS 

I.  1.  If  from  the  same  point  in  a  line  any  number 
of  lines  are  drawn  on  the  same  side  of  the  line,  the  sum 
of  the  successive  angles  formed  equals  two  right  angles. 

Hyp.     If    from    a   point   in  /q 

AB,  PC  and  PE  are  drawn 
on    the    same    side    of    AB,     ^ 
forming  successively  ZBPC, 
Z  CPE,  and  ZEPA,  X  P 

Cone. :  then  Z  BPC  +  Z  CPE  +  Z  EPA  =  2  rt.  A. 
Dem.     If  PB  rotates  to  PA,  it  generates  a  straight  Z. 

(Def.  of  a  straight  Z.) 
In  this  process  of  rotation,  PB  generates  successively  Z  BPC, 
ZCPE,?iXi^ZEPA. 

.'.  Z  BPC  +  Z  CPE  +  Z  EPA  =  2  rt.  A.       (Ax.  4.) 

Q.E.D. 

I.  1.  ScH.  If  the  angles  are  on  both  sides  of  the  line,  their 
sum  equals  four  right  angles. 

Ex.  1.  Z  J.  is  the  supplement  of  Z-B.  They  bear  to  each  other  the 
relation  of  4  to  7.     What  part  of  a  rt,  Z  is  each  ? 

Ex.  2.  If  Z.  AMC,  adjacent  to  Z  CMB,  is  four  thirds  of  a  rt.  Z,  and 
Z  CMB  is  four  fifths  of  a  rt.  Z,  are  their  exterior  sides  in  the  same  straight 
line  ? 

Ex.  3.  Three  successive  angles  about  a  point  on  one  side  of  a  straight 
line  are  in  the  ratio  of  the  numbers  2,  3,  and  5.  What  is  the  value  of 
each  angle  ? 

19 


20  THE  ELEMENTS  OF   GEOMETRY 

I.  2.  If  two  angles  are  adjacent  and  have  their  ex- 
terior sides  in  the  same  straight  line,  they  are  supple- 
mental. 

Hyp.  UZAPCandZCPB 
are  adjacent,  and  if  PB  and 
PA  are  in  the  same  straight 
line. 


>  A  P  B 

Cone. :  then        Z  CPA  +  Z  CPB  =  2  rt.  A. 

Dem.     If  PB  rotates  to  PAy  it  generates  a  straight  Z. 

(Def.  of  a  straight  Z.) 

In  this  process  of  rotation,  PB  generates  successively  Z  BPC 
and  Z  OPA 

.-.  Z  ^PC  +  Z  CP^  =  2  rt.  A.  (Ax.  4.) 

Q.E.D. 

Note.  —  The  above  proposition,  which  is  a  slight  modification  of 
Theorem  1,  is  introduced  to  assist  the  pupil  to  a  clear  statement  and 
understanding  of  its  important  converse,  which  directly  succeeds. 

I.  3.  If  tivo  angles  are  adjacent  aiid  supplemental, 
their  exterior  sides  form,  the  same  straight  line. 

Hyp.      If   A  ACL  and      x' 
ALCB  are  supplemental 
and  adjacent,  J"  ^,  n  t 

Cone:  then  AG  is  in  same  straight  line  with  CB. 

Dem.  If  CB  is  not  in  the  same  straight  line  with  AC,  draw 
CT  that  is  in  the  same  straight  line. 

Then                    A  LCT  ■\-AACL  =  2it  A.  (I.  1.) 

But                       ALCB-\-AACL  =  2xtA.  (Hyp.) 

.•   ALCT=ALCB.  (Ax.  1.) 

.-.  CT  falls  on  CB.  (Ax.  7.) 

But  CT  was  drawn  in  the  same  straight  line  with  AC. 
.'.  CB  and  CA  are  in  the  same  straight  line. 

Q.E.D 


I.     ADJACENT  AND  VERTICAL  ANGLES 


21 


I.  4.  If  two  straight  lines  intersect,  the  vertical  aiigles 
formed  are  equal. 

A 

Ryp.     If  AB  intersects 
CE, 


Cone. :  then  Z  APC  =  Z  EPB. 

Dem.  Z  CPA  +  Z  APE  =  2  rt.  A 

Z  ^PS  +  Z  ^P£;  =  2  rt.  A. 


(1.2.) 
(I.  2.) 
Z  CP^  +  Z  ^P^  =  Z  ^P5  +  Z  APE.  (Ax.  1.) 


.-.  Z  CPA  =  Z  ^P^. 


(Ax.  2.) 

Q.E.D. 


The  reference  number  only  is  given  when  the  reason  theorem  belongs 
to  the  same  group  as  the  theorem  in  course  of  demonstration. 


Ex.  4.  The  hands  of  a  clock  at  three  o'clock  form  an  angle  equal  to  a 
rt.  Z.  This  angle  will  fit  the  space  about  the  pivot  of  the  hands  exactly 
four  times.  How  many  times  is  two  thirds  of  a  rt.  Z  contained  in  a  peri- 
gon  ?    Four  thirds  of  a  rt.  Z  ?    Four  fifths  of  a  rt.  Z  ? 

Ex.  5.  The  bisectors  of  two  supplemental  adjacent  angles  form 
a  rt.  Z. 

Ex.  6.   If  the  bisectors  of  two  adjacent  A  are  ±  to 
each  other,  the  A  are  supplementary. 

Ex.  7.   The  bisector  of  an  angle  is,  when  produced, 
the  bisector  of  its  vertical  angle. 

Ex.  8.  The  bisectors  of  a  pair  of  vertical  angles  form 
the  same  straight  line. 

Ex.  9.  The  bisectors  of  two  pairs  of  vertical  angles 
are  perpendicular  to  each  other. 

Ex.  10,  If  the  sides  of  ZL  CM  are  perpendicular  to  the 
sides  of  ZACB,  prove  that  the  angles  are  supplemental. 

Ex.  11.  If  through  a  point,  A^  four  straight  lines, 
AB,  AC,  AE,  AF,  are  drawn  so  that  ZBAC=  ZEAF, 
and  ZBAF=  ZCAE,  then  FAC  and  BAE  are  straight  lines. 

Ex.  12.    What  relation  does  Ex.  11  bear  to  I,  4  ? 


22  THE  ELEMENTS  OF  QEOMETKY 


I.    SUMMARY  OF  PROPOSITIONS  IN  GROUP  ON  ADJA- 
CENT AND   VERTICAL   ANGLES 

1.  If  from  the  same  point  in  a  line  any  number  of 
lines  are  draion  on  the  same  side  of  the  line,  the  sum 
of  the  successive  angles  formed  equals  two  right  angles. 

ScH.  If  the  angles  are  on  both  sides  of  the  line,  their  sum 
equals  four  right  angles. 

2.  If  iivo  angles  are  adjacent  and  have  their  exterior 
sides  in  the  same  straight  line,  they  are  supj^lemental. 

3.  If  tivo  angles  are  adjacent  and  supplemental,  their 
exterior  sides  form  the  same  straight  line. 

4.  If  two  straight  lines  intersect,  the  vertical  angles 
formed  are,  equal. 


11.     THE   PARALLEL   GROUP 

DEFINITIONS 

Two  lines  are  said  to  be  Parallel  when  they  are  so  situated 
that  if  cut  by  any  transversal  the  corresponding  exterior- 
interior  angles  are  equal. 

Two  Lines  Perpendicular  to  a  Third.  Direct  inferences  from 
the  definition  of  parallels : 

(1)  If  two  lines  are  perpendicular  to  a  third,   they   are 

parallel. 

(2)  A  line  perpendicular  to  one  of  two  parallels  is   per- 

pendicular to  the  other. 

11.  1.  If  two  parallels  are  crossed  hy  a  third  line,  the 
alternate  interior  angles  are  equal, 

F/ 
Hyp.   If  AB  and  CE 

are  II, 
and  are  crossed  by  the 
transversal  FK,  c  H/ 

Cone. :  then  AAGH^A  GHE, 

Dem.  Z  FGB  =  Z  GHE.  (Def.  of  lis.) 

ZFGB=^ZAGH. 
[If  two  straight  lines  intersect,  the  vert. /4  formed  are  =.]  (1. 4.) 
.-.  Z  AQH  =  Z  GHE,  (Ax.  1.) 

QE.D. 


24  THE  ELEMENTS  OF  GEOMETRY 

II.  la.  If  two  parallels  are  crossed  by  a  third  line, 
the  alternate  exterior  angles  are  equal. 

Hyp.   If  AB  and  CE 

are   II, 
and  are  crossed  by  the 
transversal  KF 

K' 
Cone. ;  then  Z  AGF  =  Z  EHK 

Dem.  ZAGF  =  ZBOH. 

[If  two  straight  lines  intersect,  the  vert.  A  formed  are  =.]  (1. 4.) 

ZBGH^^ZGHC.  (ILL) 

ZGHC  =  ZEHK.  (L4.) 

.-.  Z  AGF  =  Z  EHK  (Ax.  1.) 

Q.E.D. 

II.  2.  If  two  parallels  are  crossed  hy  a  third  line,  the 
corresponding  interior  angles  are  supplermntaL 

Ff 

Hyp.    If  AB  and  CE 

are  II, 
and  are  crossed  by  the 
transversal  KFy 

K' 
Cone. :  then      Z  BGII+  Z  GHE  =  2  rt.  A. 

Dem.  Z  BGH  +  Z  BGF  =  2  rt.  A.  (1.) 

[If  from  the  same  point  in  a  line  any  number  of  lines  are 

drawn  on  the  same  side  of  the  line,  the  sum  of  the  successive 

A  formed  =  2  rt.  A.~\  (I.  1.) 

Z  BGF  =  Z  GHE.  (Def .  of  lU.) 

Substituting  Z  GHE  for  its  equal  Z  BGFj  we  have,  from  (1), 

Z  BGH+  Z  GHE  =  2  rt.  A. 

Q.E.D. 


II.   PARALLELS  25 

II.  2  a.  If  two  parallels  are  crossed  hy  a  third  line, 
the  corresponding  exterior  angles  are  supplemental. 

Hyp.     If  AB  and  CE 

are  II, 
and  are  crossed  by  the 
transversal  KFy  W 

Cone. :  then      AAGF-^Z.  CHK  =  2  rt.  A 

Dam.  ZAGF  +  ZAGH=2vt.A. 

[If  from  the  same  point  in  a  line  any  number  of  lines,  etc.] 

(I.  1.) 
Z  AGH  =  Z  CHK  (Def .  of  lis.) 

.•.  as  in  II.  2, 

Z  AGF  +  ZCHK^  2  it.  A. 

Q.E.D. 

II.  3.  If  tivo  lines  are  crossed  hy  a  third  so  as  to 
make  the  alternate  interior  angles  equal,  the  lines  are 
parallel 

Hyp.  If  TL  crosses 
AB  and  CE  so  that  the 
alt.  int.  AAVS  and 
VSE  are  equal, 


Cone. :  then  AB  II  CE. 

Dem.  ZAVS==ZTVB. 

[If  two  straight  lines  intersect,  the  vert.  A,  etc.]  (I.  4.) 

ZAVS  =  ZVSE.  (Hyp.) 

.-.  Z  TVB  =  Z  VSE.  (Ax.  1.) 

.-.  AB  II  CE.  (Def.  of  lis.) 

Q.E.D. 


26  THE  ELEMENTS  OF  GEOMETRY 

II.  3  a.  If  two  lines  are  crossed  hy  a  third  so  as  to 
make  the  alternate  exterior  angles  equal,  the  lines -are 
parallel,  ^t 

Hyp.  If  AB  and  CE 
are  crossed  by  the  trans- 
versal TL,  and  if  the  alt. 

ext.  A  TVB  and  CSL  are     "o         ^7"  F 

equal,  L/^ 

Cone. :  then  AB  II  CE. 

Dem.  Z  CSL  =  Z  VSE. 

[If  two  straight  lines  intersect,  the  vert.  Af  etc.]        (I.  4.) 
Z  CSL  =  Z  TVB.  (Hyp.) 

.-.  Z  VSE  =  Z  TFB.  (Ax.  1.) 

.-.  AB  II  Cii;.  (Dei.  of  lis.) 

Q.E.D. 

II.  4.  If  two  lines  are  crossed  hy  a  third  so  as  to 
make 

(1)  the  corresponding  interior  angles  supplemental,  or 

(2)  the  co7Tesponding  exterior  angles  supp)lemental, 
the  lines  are  parallel.  t 

Hyp.  (1)  If  the 
corr.  int.  ABVS 
and  VSE  are  sup- 
plemental, 

Cone. :  then  AB  11  CE. 

Dem.  Z.BVS  +  ZVSE  =  2vt.A.  (Hyp.) 

Z.BVS  +  ATVB  =  2v%.  A. 

[If  from  the  same  point  in  a  line  any  number  of  lines,  etc]  (T.  1.) 

.-.  Z  VSE  =  Z  TVB.  (Ax.  1.) 

.-.  AB  II  CE.  (Def.  of  lis.) 

Q.E.D. 


0         sx  JS 


n.     PARALLELS  27 

Hyp.   (2)  If  the  corr.  ext.  A  TVB  and  LSE  are  supplemental, 

Cone. :  then  AB  II  CE. 

Dem.     Similar  to  that  of  II.  4  (1). 
(Let  the  pupil  supply  the  proof.) 

II.  5.  (1)  If  two  arigles  have  their  sides  respectively 
parallel,  they  are  equal,  if  of  the  same  kind. 

Fig.  I.  Fig.  II. 


Hyp.     (1)     If 

Z.A  and  Z  B  are 
of  the  same  kind, 
and   if  ACWBF 

A\ 

\ 

\ 

\ 

E 

E             Ji\ 

\ 

\ 

\ 

K\  H 

and  AE  11  BH, 

\  ''S 

H 

Cone. :  then        ZA  =  ZB  in  Fig.  I  and  Fig.  II. 

Dem.     Fig.  I.    Extend  AG  to  cross  BH  at  K. 

ZA  =  Z  HKL.  (Def .  of  lis.) 

AB  =  A  HKL.  (Def.  of  lis.) 

,\AA  =  Z.B.  (Ax.  1.) 

Q.E.D. 

Note.  —  In  Fig.  I  the  sides  of  the  angles  extend  in  the  same  direction 
from  the  vertices. 

Dem.     Fig.  II.    Extend  AC  to  cross  BH  at  K, 

AA  =  AAKH.  (11.  1.) 

Z.B  =  Z.AKH,  (Def.  of  lis.) 

.-.  Z  ^  =  Z  5.  (Ax.  1.) 

Q.E.D. 

Note.  —  In  Fig.  II  the  sides  of  the  angles  extend  in  opposite  directions 
from  the  vertices. 


28  THE  ELEMENTS  OF  GEOMETRY 

II.  5.  (2)  If  two  angles  have  their  sides  respectively 
parallel,  they  are  8up>plemental,  if  of  different  kinds. 


Hyp.  U  ZAsLudZB 
are  of  different  kinds, 
and  if  AF  W  BC  and 
AE  II  HB, 


Cone. :  then       ZAis  supplemental  to  Z  B. 

Dem.     Extend  AE  to  intersect  BC  at  T. 

ZAis  supplemental  to  Z  ATC.  (II.  2.) 

ZATC  =  ZB.  (Def.  of  lis.) 

.'.  ZAis  supplemental  to  Z  B.  (Ax.  1.) 

Q.E.D. 

Note.  — Two  sides  ot  ZA  and  Z.B  extend  in  the  same  direction  from 
the  vertices  A  and  B^  while  the  other  two  sides,  BH  and  AE^  extend  in 
opposite  directions  from  A  and  B. 

The  theorem  may  therefore  be  stated  thus  : 

If  two  angles  have  their  sides  respectively  parallel,  and  two  extend  in 
the  same,  two  in  the  opposite  directions  from  their  vertices,  they  are 
supplemental. 

Ex.  1.  If  two  parallels  be  crossed  by  a  transversal,  and  any  angle  is 
a  right  angle,  what  is  the  value  of  each  of  the  others  ? 

Ex.  2.    Lines  II  to  the  same  line  are  ||  to  each  other. 

Ex.  3.    Show  that  Theorem  .3  is  the  converse  of  Theorem  1. 

Ex.  4.  Prove  that  a  line  parallel  to  the  base  of  a  triangle  cuts 
off  a  A  whose  angles  are  respectively  equal  to  the  angles  of  the 
original  A. 

Ex.  5.   The  bisectors  of  a  pair  of  alt.  int.  A  of  parallels  are  paralleL 
.  Ex.  6.   The  bisectors  of  a  pair  of  corresponding  interior 
angles  are  perpendicular  to  each  other, 

Ex.  7.   The  parallel  to  the  base  of  an  isoangular  triangle 
cuts  off  another  isoangular  triangle. 


n.     PARALLELS  29 


n.   SUMMARY  OP  PROPOSITIONS  IN  PARALLEL  GROUP 

1.  If  two  parallels  are  crossed  by  a  third  line,  the 
alternate  interior  angles  are  equal, 

a  If  two  parallels  are  crossed  by  a  third  line,  the 
alternate  exterior  angles  are  equal. 

2.  If  two  parallels  are  crossed  by  a  third  line,  the 
corresponding  interior  angles  are  supplemental. 

a  If  tivo  parallels  are  crossed  by  a  third  line, 
the  corresponding  exterior  angles  are  sup- 
plemental. 

3.  If  two  lines  are  crossed  by  a  third  so  as  to  make 
the  alternate  i?iterior  angles  equal,  the  lines  are  parallel. 

a  If  two  lines  are  crossed  by  a  third  so  as  to 
make  the  alternate  exterior  angles  equal,  the 
lines  are  parallel. 

4.  If  two  lines  are  crossed  by  a  third  so  as  to  make : 

(1)  the  corresponding  interior  angles  supplemental, 
or 

(2)  the  corresponding  exterior  angles  supplemental, 
the  lines  are  parallel. 

5.  (1)  If  two  angles  have  their  sides  respectively  paral- 
lel, they  are  equal,  if  of  the  same  kind. 

(2)  If  two  angles  have  their  sides  respectively  paral- 
lel, they  are  supplemental,  if  of  different  kinds. 


III.    THE  (2w-4)  RIGHT  ANGLES   GROUP 
Briefly:  The  (2^  —  4)  Group 

DEFINITIONS 

A  triangle  is  a  figure  formed  by  the  intersection  of  three 
lines  not  passing  through  the  same  point. 

If  no  two  sides  of  a  triangle  be  equal,  the  triangle  is  said  to 
be  Scalene  (limping). 

If  two  sides  of  a  triangle  be  equal,  the  triangle  is  said  to  be 
Isosceles. 

If  two  angles  of  a  triangle  be  equal,  the  triangle  is  said  to  be 
Isoangular. 

If  three  sides  be  equal,  the  triangle  is  said  to  be  Equilateral. 

If  three  angles  be  equal,  the  triangle  is  said  to  be  Equiangular. 

If  three  angles  be  acute,  the  triangle  is  called  an  Acute  Tri- 
angle. 

If  one  angle  be  obtuse,  the  triangle  is  called  an  Obtuse  Tri- 
angle. 

If  one  angle  be  right,  the  triangle  is  called  a  Right  Triangle. 

In  a  right  triangle  the  side  opposite  the  right  angle  is  the 
Hypotenuse. 

An  Altitude  of  a  triangle  is  a  perpendicular  from  a  vertex  to 
the  opposite  side.     This  side  is  called  the  Base. 

A  Median  of  a  triangle  is  a  line  from  a  vertex  to  the  middle 
point  of  the  opposite  side. 

The  Vertex  Angle  of  a  triangle  is  the  angle  opposite  the  base. 

The  Exterior  Angle  of  a  triangle  is  the  angle  between  one 
side  and  a  second  side  produced. 

30 


III.     THE    (271-4)    RIGHT  ANGLES  GROUP  31 

III.  1.  If  a  figure  is  a  triangle,  the  sum  of  the  inte" 
rior  angles  equals  two  right  angles. 

Hyp.     If  /X  /^ 

the     figure  /  N.  / 

ABC   is    a  /  \^         / 

triangle,  /  \.  / 

Z Jii 1/ 

A  £  £ 

Cone. :  then         ZA^/LC  +  Z  ABC  =  2  rt.  A, 

Dam.     Produce  AB  to  any  point  E. 

Draw  BF  ||  AC, 

Then  ZA  =  Z  FBE.  (Del  of  lie.) 

ZC^ZFBC. 

[If  two  lis  are  crossed  by  a  third  line,  the  alt.  int.  A,  etc.]  (II.  1.) 

But  Z  FBE  +  Z  FBC  +  Z  ABC  =  2  vt  A. 

[If  from  the  same  point  in  a  line  any  number  of  lines  are 
drawn  on  the  same  side  of  the  line,  the  sum,  etc.]  (I.  1.) 

.-.  ZA  +  ZC  +  ZABC  =  2i±Z. 

Q.E.D. 

III.  1  a.  One  interior  angle  of  a  triangle  is  the 
supplement  of  the  sum  of  the  second  and  third  angles. 

III.  1  h.  In  a  right  triangle,  either  acute  angle  is 
the  complement  of  the  other  acute  angle. 

Ex.  1.  In  a  triangle  Za  =  2Zb,  Zb  =  SZc.  What  is  the  value  of 
Z  rt,  Z  6,  and  Zc? 

Ex.  2.  In  a  triangle  Za  +  Zb  =  ^rtZ,  Za-Zb=i  it.  Z.  What 
is  the  value  of  Z a,  Z  b,  and  Zc? 

Ex.  3.    What  is.  the  value  of  each  acute  Z  in  an  isoangular  right  A  ? 

Ex.  4.  If  2  A  of  one  A  are  equal  respectively  to  2  zl  of 
another,  the  3d  A  are  equal. 

Ex.  5.  The  vertex  Z  of  an  isoangular  triangle  is  f  rt.  Z. 
Find  the  value  of  the  Z  between  the  base  and  an  altitude  on 
one  leg. 


THE  ELEMENTS  OF  GEOMETRY 


III.  2.  Any  exterior  angle  of  a  triangle  equals  the 
sum  of  the  two  non-adjacent  interior  angles. 

Hyp.      If 

the  figure 
ABC  is  a 
triangle,  ^  ^ 

Cone. :  then  Z  BGE  =  ZA  +  Z  B. 

Dam.  Z  ECB  +  Z  BOA  =  2  rt.  A 

[If  from  the  same  point  in  a  line  any  number  of  lines  are 
drawn  on  the  same  side  of  the  line,  the  sum,  etc.]  (I.  1.) 

ZA  +  ZB->rZACB  =  2vt.A.  (III.  1.) 

.-.  ZBCE  +  ZBCA  =  ZA  +  ZB-\-ZACB.     (Ax.  1.) 

.-.  Z  BCE  =  ZA  +  ZB.  (Ax.  2.) 

Q.E.D. 

III.  2.  SoH.  The  exterior  angle  of  a  triangle  is  greater  than 
either  of  the  non-adjacent  interior  angles. 

III.  2  a.  The  exterior  vertex  angle  of  an  isoangular 
triangle  equals  twice  either  interior  base  angle. 

III.  S.  If  a  figure  is  a  polygon  of  n  sides,  the  sum 
of  the  interior  angles,  equals  (2  ti  — 4)  right  angles. 


"Ryp.  If  ABO  •••  (?  is  a  polygon  of  n  sides, 

Cone:  thenZA  +  ZB-^ZC,  etc.  =  (2  w  -  4)  rt.  A 


III.     THE   (2n-4)    RIGHT  ANGLES   GROUP  33 

Dem.  From  any  point  0  within  the  polygon  draw  the  joins, 
OA,  OB,  OC,  etc. 

We  til  us  obtain  as  many  triangles  as  there  are  sides,  namely,  n. 
The  sum  of  the  interior  angles  of  each  A  =  2  rt.  A.       (III.  1.) 

.-.  the  sura  of  the  int.  A  oi  7i  A  =  n  x2  rt.  A,  or  2n  rt.  A. 

But  the  base  angles  only  of  these  triangles  compose  the  angles 
of  the  polygon. 

.•.  from  the  2 nit.  A  we  must  subtract  the  sum  of  the  angles 
about  O,  or  4  rt.  A.  (Sch.  to  I.  1.) 

.-.  Z  A-\-AB-\-A  C,  etc.  =2  n  rt.  ^-4 rt.  A,  or  (2  ri-4)  rt.  A. 

Q.E.D. 

III.  3  a.  In  a  regular  polygon  each  interior  angle 
equals  right  angles,      ^        ^ 


Hyp.  If  a  polygon 
ABC  "•  iy/ of  w  sides 
is  regular, 


Cone. :  then  any  interior  angle,  as  A,  =  — rt.  A. 

n 

Dem.     AA  =  AB  =  AC=  etc.     (Def.  of  regular  polygon.) 
.-.  AA  +  AB  +  AC+'"  =  nAA={2n-^)vt.A.  (III.  3.) 

.-.  AA  =  ^  of  (2n-4)rt.zi  =  ?A=irt.  A 
n  n 

Q.E.D. 

Ex.  6.    The  sum  of  the  exterior  oblique  angles 
of  a  right  triangle  equals  3  rt.  A. 

Ex.  7.    What  is  the  sum  of  the  interior  angles   __ 
in  a  4-8ide  ?  a  pentagon  ?  a  hexagon?  an  octagon ? 

Ex.  8.    What  is  the  value  of  each  interior  angle  in  the  above  if  each 
polygon  is  equiangular  ? 

Ex.  9.  How  many  sides  has  the  regular  polygon  one  of  whose  interior 
angles  is  f  rt.  Z  ?  f  rt.  Z  ?  |  rt.  Z  ?  |  rt.  Z  ? 


34  THE  ELEMENTS  OF  GEOMETRY 

ni.     SUMMARY   OF   PROPOSITIONS   IN  THE   (2/} -4) 
QROUP 

1.  If  a  figure,  is  a  triangle,  the  sum  of  the  interior 
angles  equals  two  right  angles. 

a  One  interior  angle  of  a  triangle  is  the  supple- 
ment of  the  sum  of  the  second  and  third 
angles. 

h  In  a  right  triangle,  either  acute  angle  is  the 
complement  of  the  other  acute  angle. 

2.  Any  exterior  angle  of  a  triangle  equals  the  sum 
of  the  two  non-adjacent  interior  angles. 

ScH.  The  exterior  angle  of  a  triangle  is  greater  than  either 
of  the  non-adjacent  interior  angles. 

a  The  exterior  vertex  angle  of  an  isoangular  tri- 
angle equals  twice  either  interior  hose  angle. 

Z.  If  a  figure  is  a  polygon  of  n  sides,  the  sum  of  the 
interior  angles  equals  (2  ?i  — 4)  right  angles. 

a   In  a  regular  polygon  each  interior  angle  equals 

right  angles. 

Ex  10.  Prove  that  the  sum  of  the  ext.  zi  of  a  polygon  formed  by  pro- 
ducing the  sides  in  order  at  all  the  vertices  in  succession  equals  4  rt.  A. 

Ex.  11.  What  polygon  has  the  sum  of  the  interior  angles  equal  to  three 
times  the  sum  of  the  exterior  angles  ?    One  half  the  sum  of  the  ext.  A  ? 

Ex.  12.  If  2  A  have  their  sides  respectively  ±,  they  are  equal  if  of  the 
same  kind,  and  supplemental  if  of  different  kinds. 

Ex.  13.  Can  the  plane  space  about  a  point  be  filled  without  overlapping 
by  equiangular  triangles  ?  By  regular  pentagons  ?  By  regular  octagons 
and  squares  ?    By  regular  dodecagons  and  equilateral  triangles  ? 

Ex.  14.  If  in  a  triangle  the  altitudes  Irom  the  extremities  of  the  base 
be  drawn  to  the  two  sides,  prove  that  the  angle  formed  by  them  is  the 
supplement  of  the  vertex  angle. 


IV.   GROUP   ON   ISOSCELES  AND   SCALENE 
TRIANGLES 

PROPOSITIONS 

The  Isosceles  Triangle 
IV.  1.  If  a  triangle  is  isosceles,  it  is  isoangular. 


Hyp.     If 

A  ABC    is 
isosceles; 
that    is    if* 
a  =6, 


Cone. :  then 


ZA  =  ZB. 


Dam.     Draw  CT  bisecting  Z  ACB. 
On  CT  as  an  axis  fold  over  A  CTB  to  plane  of  A  CTA. 
Side  a  will  fall  on  side  h,  '.•  Z  BCT=  Z  ACThy  const.  (Ax.  7.) 
B  will  fall  on  A,  •/  side  a  =  side  b  by  hyp. 


TB  coincides  with  TA. 
.-.  ZA  =  ZB. 


(Ax.  6.) 
Q.E.D. 


Note. — It  will  assist  the  student  to  remember  the  sequence  of  the 
theorems  in  Group  IV  if  he  observes  that  in  the  hypothesis  of  both 
IV.  1.  and  IV.  4.  are  given  the  relations  between  the  two  sides  of  the  tri- 
angle ;  in  the  conclusion,  the  relations  between  the  angles  opposite  those 
sides. 

35 


86  THE  ELEMENTS  OF  GEOMETRY 

IV.  1  a.  If  the  vertex  angle  of  an  isosceles  triangle 
is  bisected,  the  bisector  is  identical  with 

(1)  the  altitude  to  the  base, 

(2)  the  median  to  the  base. 

Hyp.  1.  If  A  ABC  is  isosceles,  and  if  CT  bisects  ZACB, 
Cone. :  then  CT  is  identical  with  the  altitude  from  C  to  AB. 
Hyp.  2.  If  A  ABC  is  isosceles,  and  if  CT  bisects  Z  ACB, 
Cone. :  then  CT  is  identical  with  the  median  from  C  to  AB. 
ScH.  In  an  isosceles  triangle,  the  altitude  to  the  base  is 
identical  with  the  median  to  the  base.  (Ax.  10.) 

IV.  2.  If  a  triangle  is  isoangular,  it  is  isosceles. 


0 


Hyp.     If 
A  ABC    is 

isoangular ; 
that  is,  if 
ZA  =  AB, 


Cone.:  then 

Dem.    If  CB  does  not  equal  CA,  draw  CB'  that  does. 

In  other  words,  suppose  AB'C  is  an  isosceles  triangle  of 
which  CA  and  CB'  are  the  equal  sides. 

Then  Z  CB'A  =  Z  A.  (IV.  1.) 

But  ZB  =  ZA.  (Hyp.) 

.-.  Z  CB'A  =  ZB.  (Ax.  1.) 

.'.  as  CB  and  CB'  have  C  in  common,  and  as  both  make  the 
same  angle  with  AB,  it  follows  that  C^  must  coincide  with 
CB'.  (Ax.  7.) 

But  CB'  was  drawn  equal  to  CA. 

.-.  CB=CA. 

Q.E.D 


IV.    ISOSCELES  AND   SCALExNE    IRIANGLES  37 

IV.  3.  If  the  altitude  of  a  triangle  bisects  the  vertex 
angle  J  the  triangle  is  isosceles. 

Hyp.  If, 

in  A  ABG, 
CT±AB, 
and  if 
Cr  bisects 
ZACB, 

A 
Cone. :  then  A  AGE  is  isosceles, 

Dem.     On  (72'  as  an  axis  revolve  A II  to  plane  of  A  I. 

CB  will  fall  on  CA,  -.-  ZBCT=ZACT  by  hypothesis.  (Ax.  7.) 

B  will  fall  on  b,  or  its  prolongation. 

TB  will  fall  on  TA,  •.•  Z  GTB  and  ZGTA  are  right  angles 

by  hypothesis.  (Ax.  7.) 

B  falls  on  TA,  or  its  prolongation. 

As  B  lies  in  b,  and  also  in  TA,  it  must  fall  on  A,      (Ax.  5.) 

.*.  a  =  b,  and  A  AGB  is  isosceles. 

Q.E.D. 

IV.  3  a.  If  the  altitude  of  a  triangle  bisects  the  base, 

the  triangle  is  isosceles. 

Hyp.     If,  in  A  ABG,  GM±  AB,  and  if  AM=  BM, 

Cone. :  then  A  AGB  is  isosceles ;  i.e.  a  =  b. 

Dem.  is  similar  to  the  preceding.     Let  the  pupil  supply  it. 

Suggestion  for  Notation.  —  Letter  foot  of  altitude,  H]  of  median, 
M;  of  bisector  of  vertex  angle,  T. 

General  Suggestions.  Does  not  your  greatest  difficulty  in  proving  a 
theorem  lie  in  these  two  points  : 

First,  that  you  forget  the  hypothesis  and  conclusion  ? 

Second,  that  you  do  not  clearly  remember  the  definition  of  the  term 
you  are  using  ? 

It  is  most  important,  therefore,  that  you  should  know  what  an  altitude 
is ;  what  a  scalene  triangle  is ;  what  a  median  is ;  in  short,  the  exact 
meaning  of  every  term  you  use. 


88  THE   ELEMENTS  OF  GEOMETRY 

The  Scalene  Triangle 

IV.  4.  In  any  triangle  the  greater  angle  lies  opposite 
the  greater  side. 


Hyp.  If, 
in  A  ABGf 
a>b, 


Cone. :  then  Z.A>AB. 

Dem.     Lay  off  CL  =  b,  and  draw  AL. 

AZ/^C  is  isosceles.  (Def.) 

.-.  ZCAL  =  ZCLA  (IV.  1.) 

But  ZCLA>ZB. 

[The  exterior  angle  of  a  A  is  greater,  etc.]         (III.  1.  Sch.) 

.'.  its  equal,  Z  CAL  >  ZB. 

.\  all  the  more  is     Z  CAB  >  Z  B, 

Q.E.D. 

Note.  —  It  does  not  follow  that  Z  A  ia  twice  Z  B,  if  side  a  is  twice 
side  b.  The  relative  length  of  tiie  sides  of  a  triangle  as  compared  with 
the  size  of  the  angles  opposite  which  they  lie,  is  treated  in  trigonometry. 

Ex.  1.    If  a  triangle  is  equilateral,  it  is  equiangular. 

Ex.  2.   If  a  triangle  is  equiangular,  it  is  equilateral. 

Ex.  3.  If  the  vertex  angle  of  an  isosceles  triangle  is  twice  either  base 
angle,  what  is  the  value  of  each  angle  of  the  triangle  ? 

Ex.  4.  Prove  that  the  bisector  of  the  exterior  vertex  angle  of  an 
isosceles  triangle  is  parallel  to  the  base. 

Ex.  5.    State  the  converse  of  Ex.  4. 

Ex.  6.  If  the  vertex  angles  of  two 
isosceles  triangles  are  supplemental, 
the  base  angles  are  complemental. 


IV.   ISOSCELES  AND   SCALENE  TRIANGLES  39 

IV.  5.  In  any  triangle  the  greater  side  lies  opposite 
the  greater  angle. 

a 

Hyp.  If, 
in  A  ABC, 
ZA>ZB, 

Cone:  then  a>b. 

Dem.     a  must  be  >  6,  <  b,  or  =  b. 

If  a<6,  ZA<ZB.  (IV.  4.) 

If  a  =  6,  ZA  =  ZB,  (IV.  1.) 

Both  these  conclusions  are  contrary  to  the  hypothesis  (q.v.). 

.'.  a>b. 

Q.E.D. 

IV.  5  a.    In   any   right   triaiigle    the   hypotenuse   is 

greater  than  either  side. 

ScH.  A  perpendicular  is  the  shortest  distance  from  a  point 
to  a  line. 

Note.  —  Distance  from  a  point  to  a  line  is  measured  on  the  perpen- 
dicular through  the.  point. 

Ex.  7.  if  the  sides  of  a  regular  hexagon  be  produced 
until  they  meet,  prove  that  an  "equilateral  triangle  is 
formed. 

Ex.  8.  The  angle  between  the  base  of  an  isosceles 
triangle  and  the  altitude  on  one  of  the  legs  equals  one 
half  the  vertex  angle. 

Ex.  9.  li  AT  bisects  Z  GAB.  and  we  draw  X^  = 
EA,  prove  that  EL  \\  CA. 

Ex.  10.  If,  through  the  vertex  of  the  vertex  angle  S 
of  the  isosceles  A  ABC,  LC  is  drawn  ±  CA  and  CM± 
CB^  prove  Zi^Cilf  is  the  double  ofZA.     (Fig.  1,  p.  41.) 

Ex.  11..  If  ^ACB  and  LCM  (Fig.  1,  p.  41)  are  isosceles,  and  the 
sides  of  the  first  are  perpendicular  to  the  sides  of  the  second,  prove  that 
Z  L  is  the  complement  of  Z  J.. 


40 


THE  ELEMENTS  OF  GEOMETRY 


TV.  6.  In  any  triangle^  if  the  altitude  to  the  base  is 
drawn,  the  side  cutting  off  the  greater  distance  from 
the  foot  of  the  altitude  is  the  greater. 


^        ^  ia)     -         -         -  ^'  iij) 

Hyp.  If,  in  any  scalene  A  ABCf  the  altitude  CL  to  the 
base  AB  is  drawn,  and  LB  >  LA^ 

Cone. :  then  CB  >  CA. 

Dem.  (a)  If  the  altitude  falls  within  the  triangle,  lay  off 
LM^LA. 

A  ACM  is  isosceles.  (IV.  3  a.) 

.-.  /.A==Z.LMC. 

A  CLM  is  a  right  triangle,  Z  CLM  being  a  right  Z.     (Hyp.) 

.-.  Z.LMC  is  acute. 

[One  interior  angle  of  a  triangle  is  the  supplement  of  the 
sum  of  the  second  and  third  angles.]  (III.  1  a.) 

.*.  Z  GMB  is  obtuse,  being  the  supplement  of  Z  LMQ, 

Z  jB  is  acute,  being  <  Z  LilfO. 

[The  exterior  angle  of  a  triangle  is  greater  than  either  oppo- 
site  interior  angle.]  (III.  2.  Sch.) 

(IV.  5.) 


(IV.  3  a.) 


/.  CB>CM. 

CM=  CA 

/.  CB>CA. 

Or  thus :  Z  CMB  >  Z  CLM==  Z  (7Z^  >  Z5. 

.-.  CB>CM. 

CM^  CA. 

.'.  CB>CA. 

Q.E.D. 

Dem.     (b)  When  the  altitude  falls  without  the  triangle,  the 

prbof  is  the  same  as  in  (ft),  omitting  the  first  step. 


IV.    ISOSCELES   AND   SCALENE  TRIANGLES 


41 


IV.  7.  In  any  triangle,  if  the  altitude  to  the  base  is 
draivn,  the  greater  side  cuts  off  the  greater  distance 
from  the  foot  of  the  altitude. 


Hyp.    If, 

in    A  ABC, 

CL  ±  AB, 

and  if 

CB  >  CA, 

Cone. :  then 


LB  >  LA. 


Dem.     LB  must  be  <  LA,  =  LA,  or  >  LA. 
If  LB  <  LA,  then  CB  <  CA.  (IV.  6.) 

If  LB  =  LA,  then  CB  =  CA.  (IV.  3  a.) 

Each  of  these  conclusions  is  contrary  to  the  hypothesis,  viz., 
that  CB  >  CA. 

.'.  LB  must  be  >  LA. 

Q.E.D. 


Ex.  12.  In  Fig.  1,  show  that 
LM  11  AB. 

Ex.  13.  In  Fig.  2,  show  that  if 
CH  is  the  altitude  of  A  ABC,  CW 
the  altitude  of  A  CML,  CH  and 
CH'  are  in  the  same  straight  line. 


Ex.  14.  If  the  bisectors  of  Z.A  and  ZB  of  tsABC 
(Fig.  3)  intersect  in  M,  and  through  M,  EF  is  drawn  ||  to 
AB,  cutting  AC  in  E  and  BC  in  JF',  prove  that  the  AAE3f 
and  ilOF  are  isosceles. 

Ex.  15.   Prove  in  the  figure  for  Ex.  14  that  EF  =  AE  +  BF. 


Ex.  16.  Prove  that  if  a  leg  of  an  isosceles  triangle  (Fig.  4) 
is  extended  its  own  length  from  the  vertex,  the  join  of  its  ex- 
tremity with  the  extremity  of  the  base  is  perpendicular  to 
the  base. 


Fig.  4 


42  .    THE  ELEMENTS  OF  GEOMETRY 

IV.    SUMMARY   OF  PROPOSITIONS   IN   THE   GROUP   ON 
THE  TRIANGLE 

The  Isosceles  Triangle 

1.  If  a  triangle  is  isosceles,  it  is  isoangidar. 

a   If  the  vertex  angle  of  an  isosceles  triangle  is 

bisected,  the  bisector  is  identical  ivith  (1)  the 

altitude  to  the  base,  (2)  the  median  to  the  base. ' 

ScH.     In  an  isosceles  triangle,  the  altitude  to  the  base  is 
identical  with  the  median  to  the  base. 

2.  If  a  triangle  is  isoangular,  it  is  isosceles. 

3.  If  the  altitude  of  a  triangle  bisects  the  verteTi 
angle,  the  triangle  is  isosceles. 

a   If  the  altitude  of  a  triangle  bisects  the  base, 
the  triangle  is  isosceles. 

The  Scalene  Triangle 

4.  In  any  triangle  the  greater  angle  lies  ojjposite  the 
greater  side. 

a   In  any  right  triangle  the  hypotenuse  is  greater 
than  either  side. 
ScH.    A  perpendicular  is  the  shortest  distance  from  a  point 
to  a  line. 

5.  In  any  triangle  the  greater  side  lies  opposite  the 
greater  angle. 

6.  In  any  triangle,  if  the  altitude  to  the  base  is 
drawn,  the  side  cutting  off  the  greater  distance  from 
the  foot  of  the  altitude  is  the  greater. 

7.  In  any  triangle,  if  the  altitude  to  the  base  is 
drawn,  the  greater  side  cuts  off  the  greater  distance 
from  the  foot  of  the  altitude. 


IV.     ISOSCELES   AND   SCALENE   TRIANGLES 


43 


Ex.  17.    If  one  Z  of  a  A  =  the  sum  of  the  other  two,  the 
A  can  be  divided  into  isosceles  A. 

Ex.  18.   If,  in  the  isosceles  A  ABC,  Z  C  =  f  rt.  Z,  and  BE     ^/ 
is  taken  equal  to  AB,  then  the  angles  of  AEAB  equal  the 
angles  of  the  original  triangle,  and  A  ECB  is  isosceles.  ^ 


Ex.  19.   If,  in  an  isosceles  A  ABC,  from  any  point  E  in 
CB  produced,  EA  is  drawn,  then  ZBAC=  ^  ^^^  +  ZE^ 


Ex.  20.  The  angle  formed  by  the  bisectors  of  the 
interior  base  angles  of  a  triangle  equals  a  rt.  Z  +  J  the 
vertex  angle. 

Ex.  21.   The  exterior  base  angle  of  an  isosceles  triangle 
equals  the  angle  formed  by  the  bisectors  of  the  two  interior   jj; 
base  angles  :  that  is,  Z  CAE  =  ZAIB.  A 


Ex.  22.    In  AABCnCT  bisects  Z  C,  and  CH 


is  an  altitude,  prove  that  ZHCT  — 


ZA-  ZB 


Ex.  23.   Prove  that  in  a  4-side  the  sum  of  three 
interior  angles  minus  the  exterior  angle  at  the    ^ 
fourth  vertex  equals  2  rt.  A. 


H  T 


Ex.  24.  In  any  triangle  the  three  new  triangles  formed 
by  the  bisectors  of  all  the  exterior  angles  of  the  triangle  are 
mutually  equiangular. 

Ex.  25.  If,  in  an  isosceles  I\ABC,  AE  is  drawn  to  any 
point,  as  E  in  BC,  then  Z  CEA  is  greater  than  ZEAC. 

Ex.  26.    If,  in  the  isosceles  A  ^  BC,  ^  is  any  point  in  BC,    ^ 
prove  that  AE  is  greater  than  BE. 


Ex.  27.    If,  in  A  ABC,  AE  is  perpendicular  to  BC, 
then  AC+CB  is  greater  than  AE  +  EB. 
What  is  the  greatest  side  of  the  triangle  f 


.E 


44  THE  ELEMENTS  OF  GEOMETRY 

Ex.  28.  If  the  vertex  angle  of  an  iso.scele8  triangle  is  twice  the  base 
angle,  the  bisector  of  the  vertex  angle  divides  the  triangle  into  two 
isosceles  triangles. 

Ex.  29.  If,  in  an  isosceles  triangle,  either  base  angle  equals  twice  the 
vertex  angle,  the  vertex  angle  is  |  rt.  Z. 

Ex.  30.  If,  in  an  isosceles  triangle,  either  base  angle  equals  twice  the 
vertex  angle,  the  vertex  angle  is  f  rt.  Z. 

Ex.  31.  In  a  regular  pentagon,  what  is  the  value  of  the 
interior  Z.  between  two  diagonals  drawn  from  the  same 
vertex  ? 

Ex.  32.  Prove  that  these  two  diagonals  are  equal. 


V.     GROUP   ON   CONGRUENT   TRIANGLES 

PROPOSITIONS 

V.  1.  If  two  triangles  have  two  sides  aiid  the  included 
angle  of  the  first  equal  to  two  sides  a7id  the  included 
angle  of  the  second,  they  are  congruent 


Hyp.     If,   in 

A I  and  II, 

G 

AB  =  EF, 

CA  =  HE, 

and  A  A  =  ZE, 

/^\ 

A                             B 

Cone. :  then 

A  I  ^  A  II. 

Dem.     Place  A  II  on  A I  with  EF  in  coincident  superposition 
with  AB,  E  falling  on  A. 

Then^i2'willfallon^(7,because,byhyp.,ZJ5;=ZA  (Ax.7.) 
.-.  H  will  fall  on  C,  because  EH=  AC,  (Hyp.) 

.-.  HF  coincides  with  CB.  (Ax.  6.) 

.-.AI^AIi. 

Q.E.D. 

Ex.  1.   If  two  triangles  are  congruent,  the  following  homologous  lines 
{i.e.  lines  having  the  same  relative  position)  are  equal : 
(a)  the  homologous  medians  ; 
(6)  the  homologous  altitudes  ; 
(c)  the  homologous  bisectors  of  the  interior  angles. 
Ex.  2.    If  two  altitudes  of  a  triangle  are  equal,  the  triangle  is  isosceles. 
Ex.  3.    Prove  that  the  altitudes  of  an  equilateral  triangle  are  equal. 
Ex.  4.    Two  isosceles  triangles  are  congruent  if  the  vertex  angle  and  its 
bisector  in  one  are  equal  to  the  corresponding  parts  of  the  other. 

45 


46 


THE  ELEMENTS  OF  GEOMETRY 


V.  2.  If  two  triangles  have  two  angles  and  the 
included  side  of  the  first  equal  to  two  angles  and  the 
included  side  of  the  second,  they  are  congruenti 


m 


Hyp.      If, 
A  I  and  II, 
ZA=ZE, 
AB=ZF, 
and  AB=EF, 

Cone. :  then 


A  I  ^  A  II. 


Dem.     Place  A  II  on  A  I  so  that  EF  is  in  coincident  super- 
position with  AB,  E  falling  on  A. 

EH  will  fall  on  A  C,  since,  by  hyp.,  ZA  =  ZE.  (Ax.  7.) 

.-.  -fflies  on  AC,  or  its  prolongation. 

Again,  FH  will  fall  on  BC,  since,  by  hyi^.,ZF=Z  B.    (Ax.  7.) 

.-.  H  lies  on  BC,  or  its  prolongation. 

Since  //  lies  on  both  AC  and  BC,  it  must  lie  at  their  inter- 
section. (Ax.  5.) 
.-.  A  I  ^  A  11. 

Q.E.D. 

V.  3.  If  tioo  triangles  have  three  sides  of  the  one 
equal  to  three  sides  of  the  other,  they  are  congruent. 

0 
Hyp.    If,  in  A  I 
and  II, 

AO=EH, 
AB  =  EF, 

and  BC^FH, 

E  F 

Cone. :  then  A  I  =  A  II. 

Dem.     Place  A II  in  the  position  of  A  II'  with  EF  in  coinci- 
dent superposition  with  AB,  E  falling  on  A.     Draw  CH\ 
A  CAH'  is  isosceles ;  also  A  BCH'.  (Hyp.) 

.-.  Z1  =  Z2  and  Z3  =  Z4. 


V,    COHGRUENT  .TRIANGLES 


47 


[If  a  triangle  is  isosceles,  it  is  isoangular.]  (lY.  1.) 

Zl  +  Z3  =  Z2  +  ^4;  th2itis,ZACB  =  ZAH'B.    (Ax.  2.) 

.-.  A  I  ^  A  II'.  (V.  1.) 

.-.  A  I  ^  A  II.  (Ax.  1.) 

Q.E.D. 

V.  4.  If  tioo  right-  triangles  have  the  hypotenuse  and 
a  leg  of  the  one  equal  to  the  hypotenuse  and  a  leg  of 
the  other,  they  are  congruent. 

0 


1' 


Hyp.     If,   in   A   I 

/' 

andll,  C5  =  J7Fand 

/ 

CA  =  HE,   and   ZB 

/ 

/ 

and    ZF   are    right 

/ 

/ 

/ 

LI 

angles, 

/ 

/ 

A 

B        E 

F 

Cone:  then               rt. 

AI 

^  rt 

.All. 

Dem.     Place  A  I  in  position  of  A  I'  so  that  CB  is  in  coinci- 
dent superposition  with  HF,  C  falling  on  H.     EFB  is  straight. 

(I.  2.) 


[If  two  supp.  A  are  adj.,  their  ext.  sides,  etc.] 

ZB^  =  ZE. 
(An  isosceles  triangle  is  isoangular.) 

.-.  ZEHF=ZFHB'. 
A  I'  ^  A  II. 
[Two  A  and  the  included  side  of  one  equal,  etc.] 
.-.  A  I  ^  A  II. 


(IV.  1.) 
(Why  ?) 

(V.  2.) 
(Ax.  1.) 

Q.E.D. 

ScH.  Since  congruent  triangles  may  be  placed  in  coincident 
superposition,  it  follows  that  homologous  altitudes,  medians, 
angle  bisectors,  mid-joins,  and  all  other  crresponding  parts 
are  respectively  equal. 


48  THE   ELEMENTS   OF  GEOMKTRV 


V.     SUMMARY   OF   PROPOSITIONS  IN   THE   GROUP 
ON   CONGRUENT   TRIANGLES 

1.  If  two  triangles  have  two  sides  and  the  included 
angle  of  the  first  equal  to  two  sides  and  the  included 
angle  of  the  second,  they  are  congruent. 

2.  If  two  triangles  have  two  angles  and  the  included 
side  of  the  first  equal  to  two  angles  and  the  included  side 
of  the  second,  they  are  congruent. 

3.  If  two  triangles  have  three  sides  of  the  one  equal 
to  three  sides  of  the  other,  they  are  congruent. 

4.  If  two  right  triangles  have  the  hypotenuse  and  a 
leg  of  the  one  equal  to  the  hypotenuse  and  a  leg  of  the 
other,  they  are  congruent. 

ScH.  Since  congruent  triangles  may  be  placed  in  coincident 
superposition,  it  follows  that  homologous  altitudes,  medians, 
angle  bisectors,  mid-joins,  and  all  other  corresponding  parts  are 
respectively  equal. 


V.     CONGRUENT  TRIANGLES 


49 


PROBLEMS 


Prob.  I.    To  construct  a  right  triangle  having  one 
side  and  the  hyj)otemcse  given. 


Given.      The  sides 

6 

b  and  c. 

Required.     To  con- 

struct a  triangle. 

JS     \B 


OF 


Const.     Take  an  indefinite  line  EF. 

At  C,  any  point  in  EF,  erect  a  ±,  CA  =  b. 

With  ^  as  a  center  and  a  radius  =  c,  describe  an  arc  cutting 

EF,  as  in  B. 

Rt.  A  ABO  is  the  required  right  triangle. 

Q.E.F. 

Proof.     If  two  right  triangles,  etc.  (V.  4.) 

Pkob.  II.    To   construct  a   triangle^  its   three  sides 
being  given. 
Given.      The     three   , 


sides,  a,  b,  and  c. 

a — 

Required,      lo     con- 
struct the  triangle.  ^        '. 

Const.     Draw  AB=:c. 

With  ^  as  a  center  and  5  as  a  radius,  describe  an  arc. 

With  JB  as  a  center  and  a  as  a  radius,  describe  an  arc. 

Let  these  two  arcs  intersect  at  any  point  O. 

Draw  AC  and  BC. 

A  ABC  is  the  required  triangle. 


Q.E.F. 


Note.  —  No  triangle  can  be  constructed  if  c?  +  &  <  0, 


60  THE  ELEMENTS  OF  GEOMETRY 


Prob.  III. 

To  construct  an 

angle 

equal 

to  a 

given 

angle. 

^ 

Given.     Z  a. 

'/.. 

t. 

Required.  To  construct        / 

\ 

/ 

/  •- 

\ 

an  angle  ^qual 

U^a.         /_. 

\ 

A 

a' 

\ 

A 

F 

-B     A^ 

i' 

Const.     On  AB  lay  off  any  distance,  AF. 

On  AC  lay  off  any  distance,  AE. 

Draw  the  join  EF. 

Construct  a  A  whose  sides  are  AF,  FE,  and  AE,   (Prob.  II.) 

AA'C'B'^AACB, 

[Two  A  are  ^  if  three  sides  of  one  =,  etc.]  (V.  3.) 

.  Then  Z  a'  is  the  required  angle. 

Q.E.r. 

Proof.  Za'^Za.  .  (Horn.  A  oi  =  A.) 

Q.E.D. 

Prob.  IV.    TJirough  a  given  point  to  draw  a  line 
parallel  to  a  given  line. 


Given.     The  point  P  and  the 
line  AB. 


Required.   To  construct  a  line  2. M 

through  P  parallel  to  AB. 

Const.     Through  P  draw  any  line  cutting  AB,  say  at  M. 

At  P  draw  a  line  PL,  making  with  PM  an  Z  =  to  Z  PMB. 

PL  is  the  line  required. 

Q.E.F. 

Proof.  ^PMB=/.LPM.  (Const.) 

.-.  LPW  AB.  '  (11.3;) 

Q.ED. 


V.     CONGRUENT  TRIANGLES 


61 


Prob.  V.     Given  tivo  sides  and  an  angle  opposite  one 
of  them,  to  construct  the  triangle. 


Given.      The 

sides  a  and  &, 
and  t^e  angle  A 

Required.   To 

construct      the 
triangle. 


Const.     At  one  extremity  of  an  indefinite  line  AL,  construct 

an  Z  LAO  =  ZA. 

On  AOlay  oEAC=h 

With  (7  as  a  center  and  radius  equal  to  a,  describe  a  circle. 

The  point  B,  in  which  the  circle  cuts  AL,  will  be  the  third 

vertex  of  the  triangle. 

Q.E.F. 

Proof.     To  be  supplied  by  the  student. 

From  the  figures  you  will  observe  that  several  cases  arise : 

1.  a  <b;  two  triangles  satisfy  the  conditions. 

2.  a  >  6 ;  one  triangle  satisfies  the  conditions. 

3.  a  =  the  perpendicular  from  C  to  AL ;  triangle  is  a  right 
triangle. 

4.  a  <  the  ±  from  C  to  AL ;  no  A  satisfies  the  conditions. 

Verify  by  drawing  the  figures  for  each  case  not  shown. 


Ex.  5.  If  A  ^50  is  equilateral  and  ^£'=:BF  = 
CH,  show  ilmt  ^ECH,  FAE,  and  FBUkvq  con- 
gruent.    Hence  sfliow  that  A  EFH  is  equilateral. 

Ex.  6.  Show  that  in  the  A  ABC  there  are  three 
congruent  4-sides, 

Ex.  7.  If  HABC  is  equilateral  and  AE  =  CH 
=  BF,  show  that  ABE,  BCF,  and  ^Cif  are  con- 
gruent.    (See  Fig.,  p.  520 


52  THE  ELEMENTS  OF  GEOMETRY 

Ex.  8.   In  the  adjacent  figure  show  that  A  ^^If, 
nCE,  and  ACF  are  congruent.    (Data  as  in  Ex.  7.) 

Ex.  9.   Also  show   that  ^AEL,   FMB,   and 
CHQ  are  congruent. 

Ex.  10.   Hence  prove  that  A  LMQ    is    equi- 
lateral. 

Ex.  11.   Show  that  ^  ALB,  BMC,  and  CQA 
are  congruent. 

Ex.  12.   Prove  that  the  4-sides  ^iilfJ?',  ^MQJT, 
and  CQLE  are  congruent. 

Ex.  13.  Two  triangles  are  congruent  if  two  altitudes  and  a  side  to 
which  one  of  them  is  drawn  in  one  triangle  equal  the  corresponding  parts 
of  the  second  triangle. 

Ex.  14.  Two  right  triangles  are  congruent,  if  the  altitude  to  the  hypot- 
fnuse  and  the  median  to  the  hypotenuse  in  one  triangle  are  equal  to  the 
corresponding  parts  of  the  second  triangle. 

Ex.  15.  All  the  theorems  and  corollaries  concerning  congruent  triangles 
in  the  summary  have  a  condition  in  common.  "What  is  it  ?  What,  then, 
may  we  infer  must  be  one  of  the  conditions  in  order  to  prove  two  triangles 
congruent  ? 

Prove  that  two  triangles  are  congruent  if  the  following  parts  of  one  are 
equal,  and  are  similarly  situated,  to  the  corresponding  parts  of  the  other  : 

(Use  the  following  notation:  Z  A,  Z  B,  ZC;  sides  opposite  these  tri- 
angles, a,  b,  c;  altitudes  to  a,  6,  c,  are  ha,  hb,  he ;  medians  on  a,  b,  c,  are 
»ia.  wift,  mc ;  angle  bisectors  are'  ta,  tj,  tc;  r  is  the  radius  of  the  inscribed 
circle ;  Vc  is  the  radius  of  the  circumscribed  circle.) 

Ex.  16.   a,  6,  and  m^,  Z  A  being  obtuse  in  both  triangles. 

Ex.  17.    a,  6,  and  he 

Ex.  18.   c,  he,  nic- 

Ex.  19.   6,  Z  A,  he,  ZA  being  acute  in  both  triangles. 

Ex.  20.    a,  Z  B,  nie,  Z  B  being  obtuse. 

Ex.  21.  Two  angles  and  the  side  opposite  one  of  them  being  given,  to 
construct  the  triangle. 

Ex.  22:  Two  rt.  L  are  ^  if  an  acute  Z  and  a  leg  of  one,  or  an  acute  Z 
and  hypotenuse  of  one,  equal  the  corresponding  parts  of  the  other. 

Ex.  23.  Through  a  given  point  without  a  given  line,  to  draw  a  line 
making  a  given  Z  with  the  given  line. 

Ex.  24.  Construct  a  triangle,  given  the  two  base  angles  and  the  bisec- 
tor of  one  of  them. 


VI.     GROUP   ON   PARALLELOGRAMS 

DEFINITIONS 

The  Quadrilateral  or  Four-side 

A  Quadrilateral,  or  4-side,  is  a  figure  formed  by  the  intersec- 
tion of  four  lines,  no  three  of  which  pass  through  the  same 
point.     Its  alternate  angles  are  called  opposite  angles. 

A  Trapezium  is  a  4-side  upon  which  no  conditions  are  im- 
posed. 

A  Trapezoid  is  a  4-side  having  one  pair  of  parallel  sides, 
called  the  bases. 

A  Parallelogram  is  a  4-side  having  two  pairs  of  parallel  sides. 

Note.  —  A  parallelogram  in  the  above  general  form  is  called  a  Rhomboid. 

A  Rectangle  is  a  parallelogram  with  two  consecutive  angles 
equal. 

A  Rhombus  is  a  parallelogram  with  two  consecutive  sides  equal. 

A  Square  is  a  rhombus,  one  of  whose  angles  is  a  right  angle. 

The  Mid-join  of  a  Trapezoid  is  the  line  joining  the  mid-points 
of  the  non-parallel  sides. 

The  Median  of  a  Trapezoid  is  the  line  joining  the  mid-points 
of  the  parallel  sides. 

Note.  —  If  the  non-parallel  sides  of  a  trapezoid  be  produced  until  they 
meet,  the  median  of  the  trapezoid  becomes  a  part  of  the  median  of  the 
triangle  formed  by  one  base  of  the  trapezoid  and  the  non-parallel  sides 
produced.  Some  authorities  give  the  name  median  to  the  mid-join.  The 
definition  above  is  more  consistent  w^ith  the  use  of  the  term  median  in 
connection  with  triangles. 

An  Isosceles  Trapezoid  is  a  trapezoid  having  its  non-paralleJ 
sides  equal. 

The  Altitude  of  a  Trapezoid  or  of  a  Rhomboid  is  the  perpen- 
dicular distance  between  the  bases. 

A  Kite  is  a  4-side  that  has  2  pairs  of  adjacent  sides  equal. 

A  Cyclic  Four-side  is  one  whose  vertices  lie  in  a  circum- 
ference. 

53 


54  THE  ELEMENTS  OF  GEOMETRY 

PROPOSITIONS 

VI.  1,  If  a  4:'side  has  two  sets  of  opposite  sides  equal, 
it  is  a  parallelogram. 

Hyp.     If,  in  the  /^^-  '-       7^ 

4.side  A-Cy  a  =  a'  /    i  r 

and  6  =  6',  A  b  B 

Cone. :  then  a  II  a'  and  6 II 6' ;  i.e.,  the  4-side  A-E  is  a  O. 
Dem.     Draw  the  diagonal  EB. 

AI^AII. 
[Two  A  ate  ^  if  three  sides  of  the  first,  etc.]  (V.  3.) 

.-.  /.  ABE  =  Z  CEB.       (Horn.  Aoi^  A.) 
.-.  6  II  6'. 
[If  two  lines  be  crossed  by  a  transversal,  etc.]  (II.  3.) 

Similarly,  Z  EBC  =  A  AEB. 

.-.  a  II  a'. 
.•.  the  4-side  A-0  is  a,  parallelogram.  (Def.  of  a  O.) 

Q.E.D. 

VI.  la.  If  a  ^'side  is  a  parallelogram^  its  opposite 
sides  are  equal.  , 

Hyp.    If  the  4-side  Z^^^-'         /     7f 

A-C  is   a  parallelo-  /      //  ^^^^^-..^  A 


gram,  A  b 

Cone. :  then  a  =  a',  6  =  6'. 

Dem.    Draw  a  diagonal. 

AI^AII.  (V.  2.) 

.'.  a  =  a',  and  h  =  6'.     (Hom.  sides  of  ^  A.) 

Q.E.D. 

ScH.    A  diagonal  divides  a  O  into  two  congruent  triangles. 

Note. — VI.  1.  and  VI.  la.  are  "Direct"  theorems;  VI.  1'.  and  VI. 
I'a'.,  on  page  65,  are  their  "Reciprocals."  See  definition  of  reciprocal 
on  p.  11. 


7 


VI.     PARALLELOGRAMS  65 

VI.  1'.  If  a  Aside  has  tivo  sets  of  opposite  angles 
egiial,  it  is  a  parallelogravi. 

Hyp.     If,  in  the  4-side  a/ 

A-C,    ZA  =  AC    and  / 

ZB  =  ZE,  A  b  B 

Cone. :  then  a  11  a'  and  6  ll  6';  i.e.,  the  4-side  A-C  is  a  parallelo 
gram. 

Dem.  ZA==ZC,   ZB  =  /.E.  (Hyp.) 

,-.  ZA  +  Z.B=^Z.C  +  /.E.  (Ax.  2.) 

But  Z^  +  Z.8  +  Z(7  +  Z^  =  4rt.  A. 

[The  sum  of  the  interior  angles  of  a  4-side  =  4  rt.  ^.]  (III.  3.) 

.-.  Z^  +  Z5  =  2  rt.  A  (Ax.  2.) 

.-.  a  I!  a'. 

[If  two  lines  are  crossed  by  a  third,  etc.]  (II.  4  (1).) 

Similarly,  &  II  &'. 

.'.  the  4-side  A-C  is  a  parallelogram.  (Del  of  a  O.) 

Q.E.D. 

VI.  1'  a',  i/*  a  Aside  is  a  parallelo  gram,  its  opposite 
angles  are  equal 

Hyp.    If  the  4-side  ^^ yO 

A-C  is   a   parallelo-              y  A' 

gram,  / / 

Cone:  then       Z^  =  Z  C7,  and  Z5  =  ZjE?. 

Dem.  a  11  a'.  (Def.  of  a  O.) 

.*.  Z  ^  is  supplementary  to  Z  jB. 

[If  two  parallels  are  crossed  by  a  transversal,  etc.]     (II.  2.) 

Z  C  is  supplementary  to  Z  B,  (Sam©  reason.) 

A  z^=za  (Ax.  1.) 

Similarly.  Z  7?  =.  Z  i7. 

9E.©. 


66  THE  ELEMENTS  OF  GEOMETRY 

VI.  %  If  a  ^'Side  has  one  set  of  sides  both  equal  and 
parallely  it  is  a  parallelogram, 

E  h  o 


Hyp.    If  a  =  a' and  a  i  a',  y     /^^^\ 


A        '        b'  B 

Cone. :  then  6  II  6'j  i.e.,  the  4-side  is  a  parallelogram. 
Dem.    Draw  the  diagonal  EB. 

AI^AII. 

[Two  A  are  ^,  if  two  sides  and  the  included  Z  of  the  first 
=  two  sides  and  the  included  Z  of  the  second.]  (V.  1.) 

/.  Z  ABE  =  Z  BEC.     (Horn.  ^  of  ^  A  are  = .) 

.-.  h  II  h\ 

[If  two  lines  are  crossed  by  a  third  so  as  to  make  the  alter- 
nate interior  angles  equal,  the  lines  are  parallel.]  (II.  3.) 
/.  the  4-8ide  ^-C  is  a  parallelogram.                (Def.  of  a  O.) 

Q.E.D. 

VI.  ^.  If  a  ^-side  is  a  parallelogram,  the  diagonals 
bisect  each  other, 

E  & 

Hyp.     If  the  4-side  /\i^ 

A-G   is   a   parallel©-  /^^-""t^^^^ 


gram,  ■       A  ■         h'  B 

Cone. :  then  the  diagonals  AC  and  BE  mutually  bisect. 

Dem.  h  =  h'.  (VI.  1  a.) 

/.ACE  =  Z.CAB. 

[If  two  parallels  are  crossed  by  a  third  line,  etc.]        (II.  1.) 

Z  BEC  =  Z  ABE.  (Same  reason.) 

.-.  AI^AII.  (Why?) 

.-.  AM=  MG  and  BM=  ME.     (Horn,  sides  of  ^  A.) 

Q.E.D, 


VI.     PARALLELOGRAMS 


67 


VI.  4.  If  the  diagonals  of  a  parallelogram  are  equal, 
the  parallelogram  is  a  rectangle. 


Hyp.     If,  in  the 
CJA-G,  AC^BE, 


Cone:  then  AEAB^ZABG;  i.e.,  the  O  A-G  is  a  rectangle. 
Dem.  AABE^AABG. 

[If  two  A  have  three  sides  of  the  first  equal,  etc.]      (V.3.) 
.-.  Z  EAB  =  Z  CBA.       (Horn.  ^  of  ^  A.) 
But  ZEAB  +  Z.CBA=:2vtA. 

[If  two  parallels  are  crossed  by  a  third  line,  etc.]        (II  2.) 


.-.  Z^^jB  =  lrt.  Z. 

(Ax.  2.) 

ZCT^  =  lrt.  Z. 

(Ax.  2.) 

Similarly,                 Z  AEG  =  Z  BGE, 

.*.  O  A-G  is  a  rectangle. 

(Def.  of  1     1.) 

Q.E.D. 

VI.  4:  a.  If  ttvo  rectangles  have  the  base  and  altitude 
of  the  one  equal  to  the  base  and  altitude  of  the  other, 
they  are  congruent. 


Suggestion.  —  It  will  be  of  much  assistance  to  the  student  in  the  solution 
of  original  exercises,  if  he  marks  the  parts  given  in  the  hypothesis  so  as  to 
distinguish  them  from  those  mentioned  in  the  conclusion.  For  example, 
a  figure  marked  thus  might  be  interpreted  to  mean  that  if  in  a  parallelo- 
gram the  diagonals  are  equal,  the  parallelo- 
gram is  a  rectangle. 

That  is,  significant  symbols  may  be  used 
to  indicate  the  relations  given  in  the  hypoth- 
esis^ while  the  cross  is  used  to  refer  to  those 
of  the  conclusion. 


58 


THE  ELEMENTS  OF  GEOMETRY 


Ex.  1.    What  are  the  different  kinds  of  quadrilaterals,  or  4-sides  ? 

Ex.  2.    What  is  the  median  and  what  the  mid-join  of  a  trapezoid? 

Ex.  3.  Produce  the  non-parallel  sides  of  a  trapezoid  until  they  meet 
What  does  the  median  (produced)  of  tlie  trapezoid  become? 

Ex.  4.  What  is  the  converse  and  what  the  reciprocal  of  a  theorem? 
Find  an  illustration  of  each  in  Group  VI.  Find  an  illustration  of  each  in 
Group  V. 

Ex.  5.  If  through  the  vertices  of  any  triangle  lines  are  drawn  parallel 
to  the  opposite  sides,  point  out  the  three  parallelograms  formed. 

Ex.  6.  Prove  that  the  new  triangle  is  four  times  the  size  of  the 
original  triangle. 

Ex.  7.  Prove  that  if  through  the  ends  of  each  diagonal  of  a  4-8ide, 
parallels  to  the  other  diagonal  are  drawn,  a  parallelogram  is  formed  which 
is  twice  as  large  as  the  4-side. 

Ex.  8.    Prove  the  converse  of  VI.  3. 

Ex.  9.  The  bisectors  of  the  corresponding  interior  angled  of  two  parallelg 
crossed  by  a  transversal  meet  at  right  angles.    Prove. 

Ex.  10.   The  bisectors  of  the  interior  angles 
of  a  parallelogram  inclose  a  rectangle.     Prove. 

Ex.  11.    What  is  a  square  ? 

Ex.  12.   The  bisectors  of  the  interior  angles 
of  a  rectangle  inclose  a  square.     Prove. 

Ex.  13.  The  bisectors  of  the  exterior  angles  of  a  rectangle  inclose  a 
square.     Prove. 

Ex.  14.  In  what  parallelograms  do  these  bisectors  of  the  interior  angles 
coincide  with  the  diagonals  ? 

Ex.  15.  In  such  parallelograms,  what  becomes  of  the  inclosed  rectangle  ? 
See  Ex.  10. 

Ex.  16.   What  is  the  sum  of  the  interior  angles  of  a  triangle  ? 

Ex.  17.    What  is  the  sum  of  the  interior  angles  of  a  4-side  ? 

Ex.  18.   Prove  that  if  HE  and  LA  bisect  Z.E  and  Z  A,  Z  HFL  is  supple- 


ment of  ^^^ 
2         2 


(Fig.  for  Ex.  21.) 

Ex.  19.   Prove  that  if  HC  and  LB  bisect  ZC  and  Z.B,  ^HJL  is  supple- 
int  of  :^  -h  ^^.     (Fig.  for  Ex.  21.) 

Ex.  20.   Why  is 

(Fig.  for  Ex.  21.) 


^+^  the  supple- 
2  2 


mentof  ^+^? 
2         2 


Ex.  21.  Prove  that  the  four  bisectors  of 
the  interior  angles  of  a  4-side  form  a  4-side 
whose  opposite  angles  are  supplemental. 


VI.     PARALLELOGRAMS 


59 


Ex.  22.  Prove  that  the  four  bisectors  of  the  exterior  angles  of  a  4-side 
form  a  4-side  whose  opposite  angles  are  supplemental. 

Ex.  23.  Prove  that  if  two  parallelograms  have  two  sides  and  the  in- 
cluded angle  of  one  equal  to  two  sides  and  the  included  angle  of  the  other, 
they  are  congruent. 

Ex.  24.  Prove  that  if  the  diagonals  of  a  4-side  not  only  bisect  but  are 
also  perpendicular  to  each  other,  tiie  4-side  is  a  rhombus. 

Ex.  25.  Prove  that  an  isosceles  trapezoid  is  isuangular.  (Draw  per- 
pendiculars to  the  longer  base  from  the  ends  of  the  sliorier.) 

Ex  26.    Prove  that  the  diagonals  of  an  iisosceles  trapezoid  are  equal. 

Ex.  27.    Why  are  the  opp.  A  of  an  isosceles  trapezoid  supplemental  ? 

E 

Ex.  28.  If  from  the  opposite  vertices  of  a 
parallelogram  perpendiculars  are  let  fall  on  a 
diagonal,  they  are  equal. 

A  B 

Ex.  29.  If  a  line  is  drawn  parallel  to  the  base  of  an  isosceles  triangle, 
it  divides  the  triangle  into  an  isosceles  triangle  and  au  isosceles  trapezoid. 

Ex.  30.   If,  in  a  parallelogram  ^-C,  J5i^       ^  ^  ^ 

=  EH,  then  the  4-side  AFCH  is  a  parallelo- 
gram. 


Ex.  31.  If  one  of  the  diagonals  of  a  4-side  bisects 
a  pair  of  opposite  angles,  but  the  other  diagonal 
does  not,  the  4-side  is  a  kite. 

Ex.  32.  Tlie  4-side  formed  by  joining  the  ends 
of  any  two  diameters  of  a  circle  is  a  rectangle. 

Ex.  33.    The  4-side  formed  by  joining  the  ends  of 
two  perpendicular  diameters  is  a  square. 

Ex.  34.    The  bisectors   of  the  opposite  angles  of  a 
rhomboid  are  parallel. 

Ex.  35.    Any   line    through    the    mid-point    of   the 
diagonal  of  a  parallelogram  divides  the  parallelo- 
gram into  two  equal  parts. 

Ex.  36.    Express  in  terms  of  n  the  number  of 
diagonals  of  an  n-gon. 


60  THE  ELEMENTS  OF  GEOMETRY 


VI.    SUMMARY  OF  PROPOSITIONS   IN  GROUP   ON 
PARALLELOGRAMS 

\.  If  a  ^-side  has  two  sets  of  opposite  sides  equal,  it 
is  a  parallelogram. 

a   If  a  Arside  is  a  parallelogram,  its  opposite  sides 
are  equal. 

ScH.     A  diagonal  divides  a  O  into  two  congruent  triangles. 

1'.  If  a  4rside  has  two  sets  of  opposite  angles  equal, 
it  is  a  parallelogram. 
a*  If  a  4t-side  is  a  parallelogram,  its  opposite 
angles  are  equal. 

2.  If  a  Aside  has  one  set  of  sides  both  equal  and 
parallel,  it  is  a  parallelogram. 

S.Ifa  4rside  is  a  parallelogram,  the  diagonals  bisect 
each  other. 

4.  If  the  diagonals  of  a  parallelogram  are  equal,  the 
parallelogram  is  a  rectangle. 

a  If  two  rectangles  have  the  base  and  altitude  of 
the  one  equal  to  the  base  and  altitude  of  the 
other,  they  are  congruent. 


VII.     GROUP   ON   SUM   OF   LINES  AND 
MID-JOINS 

PROPOSITIONS 

VII.  1.   The  sum  of  two  sides  of  a  triangle  is  greater 
than  the  third  side. 


Hyp.     If 

ABC  is   a 
triangle, 

Cone. :  then 


h-\-a>c. 


Dem.     If  either  a  or  5  >  c,  no  proof  is  required. 
If  each  side  <  c,  draw  CL  _L  c. 
Then  AL  and  BL  are  each  ±  CL. 
Now  AL  <  h,  and  BL  <  a. 

[A  perpendicular  is  shortest  distance  from  a  point  to  a  line.] 

(IV.  4  a,  Sch.) 


.',  AL-\-  BL(=  c)  <  6  +  a.     (Preliminary  Th.  1.) 

Q.E.D. 

VII.  1  a.   The  difference  hetiveen  any  two  sides  of  a  tri- 
angle is  less  than  the  third  side.    (Use  Preliminary  Th.  3.) 


Ex.  1.  Prove  that  a  line  from  the  vertex  of  the  vertex  angle  of  an 
isosceles  triangle  to  any  point  in  the  base  is  less  than  either  of  the  legs  of 
the  triangle. 

61 


62  THE  ELEMENTS  OF  GEOMETRY 

VII.  2.  Tlie  sum  of  two  lines  drawn  from  any  point 
within  a  triangle  to  the  ends  of  one  side  is  less  than  the 
sum  of  thetxoo  other  sides  of  the  triangle, 

Px 
Hyp.    If  from  Pi,  Pi^  and 
Pjfi  are  drawn,  and  from  P, 
FA  and  PB  are  drawn,  but 
enveloped  by  PiA  and  PjB, 

A 

Cone:  then  P^A  + P^B> PA  + PB. 

Dem.     Extend  AP  to  intersect  P^B  at  F. 

P,A  +  P,F>AF.  (VII.  1.) 

BF+  PF  >  BP.  (VII.  1.) 

.-.  Pi^4-PiP+-BP+PP>-4P+PF+JBP. 

(Preliminary  Th.  1.) 
Take  away  PF  from  both  members  of  the  inequality  and  we 

have  p^A  +  P^B  >PA  +  PB.     (Preliminary  Th.  3.) 

Q.E.D. 

VII.  Z.  If  a  series  of  parallels  cut  off  equal  segments 
on  one  transversal, 

(a)  They  will  cut  off  equal  segments  on  every  trans- 
versal. \      \      \      \      X.^-^"^ 

Hyp.    If  G^TTand^O  \        \        \        \^^^^^^\ 
are    any    transversals ;    '  \         \   ^^•^'C^^-'-^'X  \ 

AG^  BHy  etc.,  a  set  of        \^,^'^i\^'"jK         \         \ 
parallels,  and  AB=BO        ''o\""l\         \         \         \ 
=  CE,  etc.,  A        b\        c\        F\        W~0 

Cone. :  then     (a)  GH=  HJ=  JK==  KS,  etc. 

Dem.     Draw  GL,  HM,  JQ,  etc.,  all  parallel  to  AO. 

The  four-sides  ABGL,  BCMH,  etc.,  are  m.         (Def.  of  O.) 
.-.  GL  =  AB,  HM=  BC,  JQ  =  CE,  etc.     (VI.  1  a.) 

Ex.  2.  Show  how  theorem  VII.  3  (a)  may  be  practically  utilized  in 
solving  the  problem  :  To  divide  a  given  line-segment  into  any  number  of 
equal  parts. 


VII,     SUM  OF  LINES  AND  MID-JOINS  63 

But                            AB  =  BC=  CE,  etc.  (Hyp.) 

.-.  GL  =  HM=JQ,  etc.  (Ax.  1.) 

Again,          Z  HGL  =  Z.  JHM=  Z  KJQ,  etc.  (Def.  of  ils.) 

And              Z  HLG  =  Z  JMH=  Z  A"Q,/,  etc.  (II.  5.) 

.-.  A  GHL  ^  A  HJM^A  JKQ,  etc.  (V.  2.) 

.:  GB=HJ=JK=KS,etG. 

Q.E.D. 

VII.  oh.  If  the  series  consists  of  three  consecutive 
parallels  termiiiating  in  the  transversals,  the  mid- 
parallel  equals  half  the  sum  of  the  other  tivo.  That 
is,  the  7nid-join  of  a  trapezoid  equals  half  the  sum  of 
the  two  parallel  sides.  .  \ 

fiyp.     If  GA,  HB,  and  \       _\-.r-rrr::^r 

JC   are    three    successive  - — qS^             "A                \ 

parallels    terminating    in  \               \               \ 

the  transversals  AO  and  \ \ \ 

GW,  ^        ^V         ^ 

Cone:  then  HB  =  i(GA-\-JC); 

that  is,  the  mid-join  of  a  trapezoid  equals  one  half  the  sum  of 

the  bases. 

Dam.  HJ^HG.  (VII.  3  a.) 

.-.  BH  is  the  mid-join  of  the  trapezoid  AC  JO, 
Draw  GL  and  HQ  W  AO. 

GA  =  LB,  CQ  =  LB  +  LH.  (VI.  1  a.) 

Ce/=  CQ  +  QJ,  and  LH=  QJ. 

(Hom.  sides  of  ^  A  are  =.) 

.-.  GA  +  CQ  +  CJ=  LB-\-LB  +  LH+CQ  +  QJ,    (Ax.  2.) 
or  GA+CJ=2LB  +  2LH, 

QJ.  %A±£i^LB  +  LH=HB. 

Q.E.D. 


64 


THE  ELEMENTS  OF  GEOMETRY 


VII.  3  c.  If  in  a  triangle  a  parallel  to  the  base  in 
drawn  through  the  7nid-point  of  one  side,  it  bisects  the 
other  side  and  equals  half  the  base. 


Hyp.     If  GCJ  is  a  A,  and  if  B  is 
the  mid-point  of  GO,  and  BH  ||  CJ, 


Cone. :  then  (1) 
and  (2) 


QH^HJ 
HB=W' 


Dem.     (1)  The  A  GCJ"  may  be  considered  to  be  the  trapezoid 
GACJoi  VII.  3  (6)  with  the  side  GA  reduced  to  a  point. 
.-.  the  Dem.  of  VII.  3  (6)  will  apply  to  VII.  3  (c). 
Proof.     Let  the  student  give  the  proof  in  full. 


Ex.  3.  If,  from  any  point  in  the  base  of  an  isosceles  triangle,  lines  be 
drawn  parallel  to  the  sides,  the  perimeter  of  the  parallelogram  formed 
equals  the  sum  of  the  two  equal  sides  of  the  isosceles  triangle. 

G 

Ex.  4.  If,  from  any  point  in  the  base  of  an  isosceles 
triangle,  perpendiculars  to  the  equal  sides  are  drawn, 
prove  that  their  sum  equals  the  altitude  from  either 
one  of  the  base  angles.     (Draw  FL  ±  BH. ) 

Ex.  5.  Prove  the  foregoing  proposition  for  the  obtuse 
isosceles  triangle.  . 

Ex.  6.  AB  of  the  isosceles  A  is  then  the  locus  of 
what  point  ? 


Ex.  7.  If,  from  any  point  within  an  equilateral  tri- 
angle, perpendiculars  be  drawn  to  the  sides,  prove  that 
the  sum  of  these  three  perpendiculars  equals  the  alti- 
tude of  the  triangle.     (Draw  LE  through  F  \\  AB.)        L, 


RQ 


Ex.  8.   Prove  that  the  mid-join  of  a  triangle  is  parallel  to  the  third 
•ide. 


VII.     SUM  OF   LINES   AND   MID-JOINS 


^5 


VII.  4.  Li  a  right  triangle  the  median  to  the  hypote- 
nuse equals  half  the  hypotenuse. 


Hyp.     If,  in  the  rt.  A  ABC,  M  is 
the  mid-point  of  the  hypotenuse  c, 

Cone:  then  OJf=?. 


Dem.     Draw  MJ  II  h. 

MJ II  b  and  =  \ 

.'.  Z  MJB  is  a  rt.  Z, 

and      •  A  CMB  is  isosceles. 

[If  the  altitude  of  a  A  bisects  the  base,  etc.] 

.-.  MB  =  CM.     But  MB  =  MA. 

.'.  CM=MA. 

.  Qj^^MB±MA^c^ 
2  2 


(VII.  3  (c).) 
(Why?) 

(IV.  3  a.) 

(Hyp.) 
(Ax.  1.) 

Q.E.D. 

VII.  4  a.   Conversely,  if  a  median  of  a  triangle  equals 
half  the  side  that  it  bisects,  the  triangle  is  right. 


Ex.9.   The  A  ABC  and   AFC  have  AC  common,    CF=CB,   and 
ZACB>ZACF.     Show  that  AB >  AF.  0 

Note  that  if  Z  FCB  be  bisected  by  CH,  and 
HF  be  drawn,  A  FCH  ^  A  HCB  (V.  1)  and 
.-.  FH=  HB.  But  ylil+  FH>AF,  whence  ^^ 
>^i^;  that  is, 

If  two  triangles  have  two  sides  of  the  first 
equal  respectively  to  two  sides  of  the  second  and 
the  included  angles  unequal,  the  third  sides  will  be  unequal ;  the  greater 
side  will  belong  to  the  triangle  having  the  greater  included  angle. 

Ex.  10.    State  and  prove  the  converse  of  Ex.  9. 


66 


THE  ELEMENTS  OF  GEOMETRY 


Ex.  11.  Prove  that  if  in  a  right  triangle  one  acute  angle  is  two  thirds 
of  a  right  angle,  the  hypotenuse  equals  twice  the  shorter  side.  (Draw 
the  median  to  the  hypotenuse.) 

Ex.  12.  Prove  that  the  joins  of  the  mid-points  of  the  adjacent  sides 
of  a  4-8ide  form  a  parallelogram.     (Draw  the  diagonals.) 

Ex.  13.  Prove  that  the  parallelogram  formed  in  Ex.  12  is  one  half  the 
orighial  4-8ide. 

Ex.  14.  Prove  that  the  mid-join  of  the  diagonals  of  a  trapezoid  is 
parallel  to  the  bases  and  equals  one  half  their  difference.  (Draw  the 
join  of  an  end  of  the  upper  base  and  an  end  of  the  mid- 
join.  Two  congruent  triangles  are  formed  with  the 
bases  and  diagonals.) 

Ex.  15.  The  mid-joins  of  the  three  sides  of  a  triangle 
divide  the  triangle  into  three  congruent  triangles. 

Ex.  16.  If  we  call  ABEC  a  cross  trapezoid,  prove 
that  AB-\-CE'\%  less  than  AC  ■\-  BE. 

Ex.  17.  In  any  4-side  the  sura  of  the  diagonals  is 
less  than  the  perimeter  and  greater  than  the  semi- 
perimeter. 

Ex.  18.  In  the  isosceles  A -4 J5C,  E  is  any  point  in  AB. 
Through  E  to  draw  FL  terminating  in  6  and  a  produced 
so  that  FE  =  EL. 

Ex.  19.  In  the  isosceles  A  ABO  prove  that  if  FL  is 
bisected  in  E  that  AF  =  BL. 

Ex.  20.  If,  in  a  parallelogram  (Fig.  1),  Mis  the  mid- 
point of  EC,  and  L  is  the  mid-point  of  AB,  show  that 
EL  !l  BM. 

Ex.  21.  If,  in  Fig.  1,  EL  intersects  CA  in  E, 
and  BM  intersects  CA  in  f,  show  that  F  is  the 
mid-point  of  CB. 


Ex.  22.   In  PMg.  1,  show  that  OF  = 


CF 
2  * 


Ex.  23.   In  Fig.  1,  BM  and  CO  are  medians 
of  A  BCE.     Prove  by  means  of  the  three  pre- 
ceding theorems  that  the  medians  of  a  triangle 
concur. 

Ex.  24.   If  two  medians  of  a  triangle  intersect 

AO 
2 
BE 
3  * 


in  0  as  in  the  adjoining  figure,  then  0M  = 

and  O^ = 4^  J  that  is,  OM = ^,  and  ^0  =- 
2  3 


HnrT.  — Draw  CL  y  BO. 


Vn.     SUM   OF  LINES   AND  MID^OINS 


67 


Ex.  25.  If,  in  A  ABO,  AG  and 
CH  are  two  altitudes  on  a  and  c, 
respectively,  and  TK  and  MK  are 
mid-perpendiculars  to  a  and  c,  prove 


that  Tir  = 


AI 


2 

Hint.  —  Draw  MF  =  ilf r  and 
JPi?  II  CH.  What  triangles  are  con- 
gruent ? 


Ex.  26.  The  mid-joins  of  the  four  halves 
of  the  two  diagonals  of  a  rectangle  form  a 
second  rectangle. 


Ex.  27.  If,  in  A  ABC,  CT  bisects  the  vertex 
angle  and  AE  and  BF  are  drawn  perpendicular  to 
CT,  then,  if  M  is  the  mid-point  of  AB,  the  join 
ME  (or  MF)  =  \{BC  -  AC). 

Ex.  28.  If  ^£' and  jBjP  are  drawn  perpendicular 
to  the  bisector  of  the  exterior  vertex  angle,  the 
join  ME  =  \{BC-^  AC). 

Ex.  29.    The  Z  QAB  =  lU  CAB  -AB). 

Ex.  30.  The  join  of  the  mid-points  of  two 
opposite  sides  of  a  4-side  and  the  join  of  the  mid- 
points of  its  diagonals  are  the  diagonals  of  a 
parallelogram. 

Ex.  31.  State  at  least  three  general  methods 
of  proving  that : 

Lines  are  parallel. 

Lines  are  equal. 

Angles  are  equal. 

Angles  are  supplemental. 


68  THE  ELEMENTS  OF  GEOMETRY 

VII.     SUMMARY   OF   PROPOSITIONS   IN   THE  GROUP 
ON   SUM   OF   LINES   AND   MID^OINS 

1.  The  sum  of  two  sides  of  a  triangle  is  greater  than 
the  third  side. 

(a)  The  difference  between  any  two  sides  of  a 
triangle  is  less  than  the  third  side. 

2.  The  sum  of  two  lines  draivn  from  any  point 
loithin  a  triangle  to  the  ends  of  one  side  is  less  than 
the  sum  of  the  two  other  sides  of  the  triangle. 

Z.  If  a  series  of  parallels  cut  off  equal  segments  on 
one  transversal, 

(a)  They  will  cut  off  equal  segments   on  every 

transversal. 

(b)  If  the  series  consists  of  three  consecutive  paral- 

lels terminating  in  the  transversals,  the  mid- 
parallel  equals  half  the  sum  of  the  other  two. 
That  is,  the  mid-join  of  a  trapezoid  equals 
half  the  sum  of  the  two  parallel  sides. 

(c)  If  in  a  triangle  a  parallel  to  the  base  is  drawn 

through  the  mid-point  of  one  side,  it  bisects 
the  other  side  and  equals  half  the  base. 

4.  Ii  a  right  triangle  the  median  to  the  hypotenuse 
eqitals  half  the  hypotenuse. 

(a)  Conversely,  if  a  median  of  a  triangle  equals 
half  the  side  that  it  bisects,  the  triangle  is 
right. 


VIII.     GROUP   ON   POINTS  — EQUIDISTANT 
AND   RANDOM 


PROPOSITIONS 

VIII.  1.  Every  point  on  the  mid-perpendicula?'  of  a 
Une-segment  is  equidistant  fi^om  the  ends  of  the  line- 
segment,  r^' 

Hyp.  If  RM  is  a  mid-perpen- 
dicular to  AB,  and  P  is  any  point 
in  RM, 

Cone. :  then 

Dem     PM  is  the  altitude  to  base  of  A  ABP.      (Def .  of  alt.) 
.*.  A  PAB  is  isosceles.  (IV.  3  a.) 


PA  =  PB. 


(Def.  of  isos.  A.) 

Q.E.D. 


What 
Ex.  1 
Ex.  2 
Ex.3 
Ex.  4 
Ex.  5 
Ex.  6 
Ex.  7 
Ex.  8 


is  the  locus  of  a  point  satisfying  the  following  conditions ; 
At  a  distance  a  from  a  given  point  Q  ? 
At  a  distance  a  from  a  given  line  AL  ? 
At  a  distance  a  from  a  given  circumference  K? 
Equidistant  from  Q  and  B  ? 

Equidistant  from  two  intersecting  lines  AL  and  BM? 
Equidistant  from  two  parallels  ? 
Equidistant  from  two  equal  circumferences  ? 
Equidistant  from  two  concentric  circumferences  ? 


Find  the  points  that  satisfy  the  following  conditions  : 
Ex.  9.   At  a  distance  a  from  Q,  and  a  distance  b  from  a  line  AL. 
Ex.  10.    At  a  distance  a  from  ^,  and  a  distance  b  from  a  O  K. 

69 


70 


THE  ELEMENTS  OF  GEOMETRY 


VIII.  2.  Every  point  loithout  the  mid-perpendicular 
of  a  line-segment  is  unequall]/  distant  from  the  ends 
of  the  line-segment.  p 


Hyp.  If  ML  is  a  mid-perpen- 
dicular to  AB,  and  if  P  is  any 
point  without  ML, 

Cone. :  then  PA  is  not  equal  to  PB. 


Dem.   Let  AP  intersect  the  mid  J_  ML  in  T, 
Draw  TB, 

TB=TA. 

TB+TP>PB. 

TB.-{-TP=TA+  TP  (Ax.  2)  =  PA, 

.'.  PA  does  not  equal  PB. 


(VIII.  1.) 
(VII.  1.) 

Q.E.D. 


ScH.  1.  The  mid-perpendicular  of  a  line-segment  is  the  locus 
of  points  equidistant  from  the  ends  of  the  line-segment. 

ScH.  2.  Two  points,  each  of  which  is  equidistant  from  the 
ends  of  a  line-segment,  determine  the  mid-perpendicular  to  the 
line. 

VIIT.  3.  Every  point  in  the  bisector  of  an  angle  id 
equidistant  from  the  sides  of  the  angle. 

jj 

Hyp.     If  ^T   is   the   bi-  ^- 

sector  of  Z  BAC,  P  any  point 
in  AT,  and  if  PF  and  PE 
are  perpendiculars  to-4B  and 
AC,  respectively, 

Cone. :  then  P  is  equidistant  from  AB  and  AC.  That  is, 
PF=PE. 


VIII.     POINTS  — EQUIDISTANT  AND   RANDOM  71 

Dem.  ZEAP=ZFAP.  (Hyp.) 

Z  EPA  =  Z  FPA.  (Ax.  1.) 

AP=AP. 

.-.  AEAP^AFAP. 

[Two  Z  and  the  included  side  of  the  first,  etc.]  (V.  2.) 

.-.  PF=  PE.         (Horn,  sides  of  ^  A.) 

Q.E.D. 

Vlli.  4.  Every  point  ivitJiout  the  bisector  of  an  angle 
IS  unequally  distant  from  the  sides  of  the  angle, 

G 
Hyp.     If  AT   is    the   bi- 
sector of  Z  BAC,  P  any  point  ^-^""^^^X  / 
without  AT,  and  if  PF  and 
PE  are  perpendiculars  to  AB 
and  AC,  respectively,  /^ 

B 
Cone. :   then  P  is  unequidistant  from  AB  and  AC.     That  is, 
PF  does  not  =  PE. 

•    Dem.     Draw  MffX^C;  also  the  join  P5". 

PH>PE. 

[A  ±  is  the  shortest  distance  from  a  point  to  a  line.] 

(IV.  4  a.  Sch.) 

[The  sum  of  two  sides  of  a  A  is  greater,  etc.]  (VII.  1.) 

MH=FM.  (VIII.  3.) 

.-.  FM+MP>PH. 

But  FM+MP=  PF.  (Ax.  4.) 

/.  PF>  PII,  which  is  greater  than  PE. 

.-.  PF>PE. 

Q.E.D. 


72.  THE  ELEMENTS  OF  GEOMETRY 

8cu.  1.  Two  points,  each  of  which  is  equidistant  from  the 
sides  of  an  angle,  determine  the  bisector  of  the  angle. 

ScH.  2.  The  bisector  of  an  angle  is  the  locus  of  all  points 
equidistant  from  the  sides  of  the  angle. 

Gen.  Sch.  Two  points  in  any  straight  line  locus  determine 
the  locus. 

Note.  —  Sch.  1.  of  VIII.  2.  affords  the  proof  of  the  solutions  of  the 
following  problems  of  the  ten  easy  exercises  in  geometrical  drawing 
(pp.  14-17): 

Prob.  I.     (a)  Bisect  a  given  line-sugment. 

(6)  Erect  a  raid-perpendicular  to  a  given  line-segment. 

Prob.  V.  Erect  a  perpendicular  to  a  given  line  at  a  given  point'  in  the 
line. 

Prob.  VI.  Draw  a  perpendicular  to  a  given  line  from  a  given  point 
without  the  line. 

Find  the  points  that  satisfy  the  following  conditions  : 

Ex.  11.    In  a  given  line  AL,  and  at  a  distance  b  from  a  given  line  Q. 

Ex.  12.    In  a  given  line  AL^  and  at  a  distance  b  from  a  OA'. 

Ex.  13.    In  a  given  circumference,  and  at  a  distance  b  from  Q. 

Ex.  14.    In  a  given  circumference,  and  at  a  distance  b  from  a  line  AL. 

Ex.  15.  In  a  given  circumference,  and  at  a  distance  b  from  a  second 
circumference. 

Ex.  16.    In  a  given  line,  and  equidistant  from  Q  and  B. 

Ex.  17.   In  a  given  circumference,  and  equidistant  from  Q  and  E. 

Ex.  18.   At  a  distance  a  from  a  line  AL,  and  equidistant  from  Q  and  B. 

Ex.  19.  At  a  distance  a  from  a  given  circumference,  and  equidistant 
from  Q  and  B. 

Ex.  20.  At  a  distance  a  from  a  given  point  Q,  and  equidistant  from 
B  and  S. 

Ex.  21.  At  a  distance  a  from  a  given  point  Q,  and  equidistant  from 
two  intersecting  lines  AL  and  AM. 

Ex.  22.  At  a  distance  a  from  a  point  Q,  and  equidistant  from  two 
parallels  AL  and  BM. 

Ex.  23.  At  a  distance  a  from  a  circumference,  and  equidistant  from 
two  intersecting  lines. 

Ex.  24.  Equidistant  from  Q  and  B,  and  also  from  two  intersecting 
lines  AL  and  BM. 


VIII.     POINTS  — EQUIDISTANT  AND   RANDOM  73 


VIII.     SUMMARY   OP  PROPOSITIONS   IN   THE  GROUP 
ON  POINTS  — EQUIDISTANT   AND   RANDOM 

1.  Every  point  on  the  mid-jjerpendicular  of  a  line- 
segment  is  equidistant  from  the  ends  of  the  line-segment. 

2.  Evey^y  point  loithout  the  mid-perpendicular  of  a 
line-segment  is  unequally  distant  from  the  ends  of  the 
line-segment. 

ScH.  1.  The  mid-perpendicular  of  a  line-segment  is  the  locus 
of  points  equidistant  from  the  ends  of  the  line-segment. 

ScH.  2.  Two  points,  each  of  which  is  equidistant  from  the 
ends  of  a  line-segment,  determine  the  mid-perpendicular  to  the 
line. 

3.  Every  point  in  the  hisector  of  an  angle  is  equidis- 
tant from  the  sides  of  the  angle. 

4.  Every  point  ivithout  the  hisector  of  an  angle  is 
unequally  distant  from  the  sides  of  the  angle. 

ScH.  1.  Two  points,  each  of  which  is  equidistant  from  the 
sides  of  an  angle,  determine  the  bisector  of  the  angle. 

ScH.  2.  The  bisector  of  an  angle  is  the  locus  of  all  points 
equidistant  from  the  sides  of  the  angle. 

Gen.  Sch.  Two  points  in  any  straight  line  locus  determine 
the  locus. 


74  THE  ELEMENTS  OF  GEOMETRY 


NINE  ILLUSTRATIONS  OP   ELEMENTARY  PRINCIPLES 
OF  LOCI 

1.  The  locus  of  a  point  at  a  given  distance  from  a  given  Point. 

2.  The  locus  of  a  point  at  a  given  distance  from  a  given  Straight  Line. 

3.  The  locus  of  a  point  at  a  given  distance  from  a  given  Circle. 

4.  The  locus  of  a  point  equidistant  from  two  given  Points. 

5.  The  locus  of  a  point  equidistant  from  two  given  Straight  Lines 
which  intersect. 

6.  Tlie  locus  of  a  point  equidistant  from  two  Parallels. 

7.  The  locus  of  a  point  equidistant  from  two  Concentric  Circles. 

8.  The  locus  of  a  point  from  which  perpendiculars  may  be  drawn  to  a 
given  straight  line, 

(a)  to  a  given  point  in  the  line  ; 

(b)  through  a  given  point  without  the  line. 

9.  The  locus  of  a  point  from  which  obliques  may  be  drawn  making  a 
given  angle  with  the  line, 

(a)  to  a  given  point  in  the  line  ; 

(6)  through  a  given  point  without  the  line. 

NINE  EXERCISES  IN  LOCI 

Because  of  the  frequent  use  of  The  questions  on  the  left  are 

the  idea  of  the  locus  in  the  subse-  given   in    familiar,   everyday  lan- 

quent  demonstrations,  it  is  of  the  guage. 

utmost  importance  that  the  pupil  The  statements  below  are  an- 

become   thoroughly  familiar  with  swers  to  the    questions    opposite, 

these  simple  yet  fundamental  no-  and  are  given  in  the  language  of 

tions  of  the  locus.  geometry. 

1.  What  line  contains  all  the  1.  The  locus  of  a  point  1  mile 
houses  that  are  1  mile  distant  from  distant  from  a  given  point  is  a  cir- 
the  city  hall  ?  cle  whose  center  is  the  given  point 

Ans.  The  circle  whose  center  is      and  whose  radius  is  1  mile, 
the  city  hall  and  whose  radius  is  The  locus  of  a  point  at  a  given 

1  mile,  distance  from  a  given  point  is  a  cir- 

What  is  necessary  to  determine  cle  whose  center  is  the  given  point 
the  size  and  position  of  a  circle  ?  and  whose  radius  is  the  given  dis- 

tance. 

Ex.     Define  a  circle  as  a  locus. 

2.  What  line  or  lines  contain  all  2.  The  locus  of  a  point  1  mile 
the  houses  1  mile  distant  from  a  distant  from  a  given  line  consists  of 
main  street  in  your  city  ?                     two  parallels  to  the  given  line,  one 

on  each  side,  1  mile  from  the  line. 


VIII.     POINTS  — EQUIDISTANT  AND  RANDOM 


75 


Note.  —  By  distance  is  meant, 
unless  otherwise  stated,  the  perpen- 
dicular distance. 

What  is  necessary  to  fix  or  de- 
termine the  position  of  a  line  ? 


3.  What  line  contains  all  the 
flower  pots  that  may  be  placed 
10  feet  from  a  circular  path  whose 
diameter  is  100  feet  ? 

The  distance  from  a  circle  is 
always  measured  on  a  radius,  or 
radius  produced. 

What  is  the  name  of  two  or  more 
circles  that  have  the  same  center  ? 


4.  What  line  contains  all  the  hy- 
drants that  may  be  placed  equidis- 
tant from  the  ends  of  a  straight 
street  ? 

What  determines  the  position  of 
this  line  ? 


5.  What  line  contains  all  the 
hydrants  that  may  be  placed  in  a 
park  so  as  to  be  equidistant  from 
two  intersecting  straight  paths  ? 

.  6.    What  line   contains  all  the 
points  that  are  equidistant  from  the 


The  locus  of  a  point  at  a  given 
distance  from  a  given  line  consists 
of  two  parallels  to  the  given  line, 
one  on  each  side,  at  a  given  distance 
from  the  line. 

Ex.  By  what  axiom  are  the 
above  lines  determined  in  position  ? 
Under  which  of  the  two  ways  of 
stating  this  axiom  does  the  determi- 
nation directly  fall  ? 

3.  The  locus  of  a  point  10  feet 
from  a  given  circle  whose  diameter 
is  100  feet,  cormsis  o/two  concentric 
circles  whose  radii,  respectively,  are 
60  feet  and  40  feet. 

The  locus  of  a  point  at  a  given 
distance  a  from  a  given  circle  whose 
radius  is  r,  consists  of  two  concentric 
circles  whose  radii,  respectively,  are 
r  -\-  a  and  r  —  a. 

Ex.  What  determines  the  posi- 
tion and  size  of  a  circle  ? 

4.  The  locus  of  a  point  equidis- 
tant .from  two  given  points  is  the 
mid-perpendicular  to  the  join  of 
the  two  points. 

Ex.  What  is  the  direction  of 
this  locus  with  reference  to  the  join 
of  the  two  points  ?  (v.  definition 
of  direction.) 

Note. — The  mid-perpendicular 
to  the  join  of  two  points  is  also  the 
locus  of  the  centers  of  circles,  any 
one  of  which  passes  through  both 
points. 

5.  The  locus  of  a  point  equidis- 
tant from  two  given  intersecting 
straight  lines  consists  of  two  lines, 
and  bisecting  the  angles  formed  by 
the  lines. 

6.  The  locus  of  a  point  equidis- 
tant from  two  parallels  is  a  line 


76 


THE  ELEMENTS  OF  GEOMETRY 


two  rails  of  a  cable  street  railroad  ? 

What  determines  the  position  of 
this  line  ? 

7.  What  lino  contains  all  the 
flower  pots  that  may  be  placed 
equidistant  from  two  concentric 
circular  paths  with  radii  of  50  feet 
and  76  feet,  respectively  ? 


8.  What  line  contains  all  the 
windows  in  a  high  building  from 
which  a  boy  may  drop  apples  into 
a  basket  standing  against  the  build- 
ing, on  the  level  sidewalk  ? 


9.  (1)  Along  what  line  should 
we  find  all  the  telegraph  poles  on 
which  wires  may  be  strung  in  north- 
east and  southwest  direction,  to 
cross  an  east  and  west  county 
road  at  the  schoolhouse  (a)  on  the 
county  road,  (6)  1  mile  from  the 
county  road  ? 

(2)  Same  question  for  a  north- 
west and  southeast  telegraph  line. 


parallel  to  them,  midway  between 

ihem. 

7.  The  locus  of  a  point  equidis- 
tant from  two  concentric  circles 
whose  radii  are,  respectively,  50  feet 
and  75  feet,  is  a  circle  concentric 
with  the  given  circles,  whose  radius 
is  62  J  feet. 

The  locus  of  a  point  equidistant 
from  two  concentric  circles  of  radii 
a  and  6  is  a  circle  concentric  with 

the  given  circles  of  radius  ^-i— . 

8.  (a)  The  locus  of  a  point  from 
which  perpendiculars  may  be  drawn 
to  a  given  line  at  a  given  point  in 
the  line  is  the  perpendicular  to  the 
line  at  the  given  point. 

(6)  The  locus  of  a  point  from 
which  perpendiculars  may  be  drawn 
to  a  given  line  through  a  given  point 
without  the  line  is  the  perpendicular 
to  the  line  from  the  point. 

9.  The  locus  of  a  point  through 
which  obliques  may  be  drawn  to  a 
given  line,  making  an  angle  equal 
to  one  half  a  right  angle  (a)  at  a 
given  point  in  the  line,  (h)  at  a 
given  point  without  the  line,  is 
the  line  through  the  given  point 
(1)  making  half  a  right  angle  with 
the  line  ;  (2)*  making  a  negative 
angle  equal  to  half  a  right  angle 
with  the  line. 


Note.  —  In  order  to  prove  that  a  locus  consists  of  a  line,  or  a  set  of 
lines,  it  is  necessary  to  show 

First,  that  every  point  on  the  line,  or  set  of  lines,  fulfills  the  given 
conditions. 

Second,  that  no  point  without  the  line,  or  set  of  lines,  does  fulfill  the 
given  conditions. 


VIII.    POINTS— EQUIDISTANT  AND   RANDOM  77 


CHART   PROBLEMS 

Note.  —  In  order  to  answer  the  questions  asked,  students  are  at  liberty 
to  change  dimensions  and  must  sometimes  for  theoretical  reasons  deal 
with  impractical  conditions. 

Let  us  assume,  for  the  purpose  of  illustration,  that  The  Pirate'' s  Chart 
gives  the  following  description  of  the  locations  of  his  buried  treasure  : 

1.  The  first  is  a  half  mile  from  an  oak,  and  at  the  same  time  is  three 
quarters  of  a  mile  from  a  chestnut. 

Locate  the  treasure.  (v.  Locus,  Ex.  1.) 

When  are  there  two  possible  locations  ?     When  none  ? 
[Draw  a  diagram  for  the  above  and  for  the  following  exercises.     In 
the  diagram  make  the  given  line  or  lines  solid,  the  loci  dotted.] 

2.  The  second  is  a  quarter  mile  from  the  shore  of  a  shallow  circular 
pond,  whose  radius  is  one  mile,  and  simultaneously  is  a  half  mile  from  a 
neighboring  straight  beach. 

Locate  the  treasure.  (v.  Locus,  Exs.  2  and  3.) 

When  may  there  be  eight  such  locations  ? 

3.  The  third  is  equidistant  from  the  oak  and  the  chestnut,  and  simul- 
taneously is  one  and  a  half  miles  distant  from  the  shore  of  a  neighboring 
circular  pond,  whose  radius  is  one  quarter  of  a  mile. 

Locate  the  treasure.  (v.  Locus,  Exs.  4  and  3.) 

When  may  there  be  four  such  locations  ? 

4.  The  fourth  is  equidistant  from  the  turnpike  and  the  valley  road,  and 
is  simultaneously  equidistant  from  the  oak  and  the  chestnut. 

Locate  the  treasure.  (v.  Locus,  Exs.  5  and  4.) 

5.  The  fifth  lies  on  a  line  making  with  the  turnpike  a  positive  two 
thirds  of  a  rt.  Z,  and  passing  through  the  oak  ;  also  on  a  line  making  with- 
the  turnpike  a  negative  one  half  of  a  rt.  Z,  and  passing  through  the  chestnut. 

Locate  the  treasure.  (v.  Locus,  Ex.  9.) 

Suppose  the  trees  were  (a)  upon  the  turnpike,  (6)  remote  from  it. 

6.  The  sixth  lies  on  a  perpendicular  to  the  turnpike  at  the  school- 
house  ;  also  on  a  line  passing  through  the  oak  and  ||  to  the  valley  road. 

Locate  the  treasure.  (v.  Locus,  Ex.  8.) 

7.  The  seventh  lies  on  a  perpendicular  through  the  oak  to  the  join  of 
the  oak  and  the  chestnut,  and  is  simultaneously  one  mile  from  the  oak. 

Locate  the  treasure.  (v.  Locus,  Exs.  8  and  1.) 

Note.  —  Descriptions  5.  6,  and  7  are  applications  of  Locus,  Exs.  8  and  0. 
These  two  exercises  should  be  given  careful  attention. 


IX.      GROUP   ON   THE   CIRCLE   AND   ITS 
RELATED    LINES 

DEFINITIONS 

A  Secant  is  a  line  cutting  a  circumference  in  two  points. 

A  Chord  is  the  join  of  any  two  points  on  a  circumference. 

The  arcs  that  have  the  same  extremities  as  a  chord  are  said 
to  be  subtended  by  the  chord.  The  greater  of  the  two  arcs  is 
called  the  major,  and  the  smaller  the  minor,  arc. 

A  Diameter  of  a  circle  or  circumference  is  a  chord  that  passes 
through  the  center. 

A  Tangent  is  a  line  that  touches  a  circle  or  circumference  in 
but  one  point. 

Two  circles  are  tangent  to  each  other  when  they  are  tangent 
to  the  same  line  at  the  same  point. 

An  Arc  is  any  part  of  a  circumference. 

A  Segment  of  a  circle  is  that  portion  of  the  circle  contained 
between  an  arc  and  the  chord  having  the  same  extremities  as 
the  arc.     This  chord  is  said  to  subtend  the  arc. 

A  Sector  of  a  circle  is  that  portion  of  the  circle  contained 
between  two  radii  and  the  arc  that  they  intercept. 

Corollaries  op  the  Definitions 

1.  Circles  with  equal  radii  are  congruent.  (See  Defini- 
tions, pp.  2,  8.) 

2.  A  line  tJiat  intersects  a  circumference  intersects  it 
in  two  points  and  no  more. 

3.  Any  diameter  of  a  circle  bisects  it. 

4.  A  tangent  may  he  considered  as  obtained  by  re- 
volving a  secant  about  either  point  of  secancy  until  the 
two  points  coincide. 

78 


IX.     THE  CIRCLE  AND   ITS  RELATED  LINES  79 

PROPOSITIONS 

IX.  1.  A  radius  perjyendicidar  to  a  chord  bisects  the 
chord  and  its  subtended  arc,  and  convei^sely, 

F 
Hyp.     If,  in 

O  K^  the  fadius 
KF  is  perpen- 
dicular to  the 
chord  AB, 


Cone. :  then      (a)  KF  bisects  the  chord  AB. 

(b)  KF  bisects  the  arc  AFB. 
Bern,   (a)  Draw  KA  and  KB. 

A  AKB  is  isosceles.         (Sides  are  radii.) 

.-.  KF  bisects  the  chord  AB. 
[In  an  isosceles  A  the  altitude  to  the  base,  etc.]  (IV.  1  a,  Sch.) 

Q.E.D. 

Dem.     (b)  On  KF  as  an  axis  revolve  the  sector  KBF  to  the 
plane  of  sector  KFA. 

B  will  fall  on  A.  (Why  ?) 

Moreover,  the  arcs  AF  and  FB  will  coincide  throughout,  as 
all  radii  are  equal. 

.-.  arc^i^=  arc^F. 

Q.E.D. 

Hyp.   Conversely,  if,  in  a  circle,  a  radius  .KFbisects  a  chord  J.5, 

Cone. :  then  /fjP  ±  chord  AB. 

Dem.  AK=  BK.  (Radii  of  same  O.) 

.•.  A  ABK  is  isosceles.  (Def.  of  isos.  A.) 

.*.  KF  is  the  altitude  to  the  base  of  isos.  A  ABK. 

[In  an  isosceles  A  the  altitude  to  the  base,  etc.]  (IV.  1  a,  Sch.) 

.-.  KF±choidAB. 

Q.E.D. 


THE  ELEMKKTS  OF  GEOMETRY 


IX.  1  a.  A  radius  2}erpe7idicular  to  a  chord  is  mid' 
^erfendiadar  to  every  chord  parallel  to  the  first      - 


Hyp.  If,  in  O  7i, 
the  radius  KF  is 
perpendicular  to  the 
chord  ^B,  and  if  CJ57 
is  any  chord  parallel 
to  the  chord  ABy 


Cone :  then  KF  is  the  mid-JL  to  the  chord  CE. 

Dem.  KFA.  AB.  (Hyp.) 

.-.  KFA.  CE.     (i)ef.  of  lis,  first  inference.) 

.-.  KF  is  the  mid-X  to  CE,  (IX.  1.) 

Q:ED. 

IX.  2.  In  the  same  circle,  or  in  equal  circles,  equal 
chords  are  equidistant  from  the  center,  and  conversely. 


Hyp.  If,  in  O  K, 
the  chord  AB  =  the 
chord  CE, 


Cone. :     then  the  ±  KH  =  the  ±  KF. 

Dem.     Draw  the  radii,  KB,  KA,  KC,  and  KE, 
A  KAB  ^  A  KCE. 

[If  two  A  have  three  sides  of  one  equal,  etc.]  (V.  3.) 

.-.  KH  =  KF.  (Sch.  to  Th.'s  ^  A,  Group  V.) 

Q.E.D. 
Hyp.   Conversely,  if  the  ±  KH=  the  l.KFm.  the  O  K, 

Cone. :  then  the  chord  AB  =  the  chord  CE. 


IX.     THE   CIRCLE   AND    ITS    RELATED   LINES  81 

Dem.     It.  A  HKA  ^  rt.  A  HKB  ^  rt.  A  FKC  ^  rt.  A  FKE. 
[Right  A  are  ^  if  a  leg  and  hypotenuse  of  one,  etc.]     (V.  4.) 

"    .-V  AH=  HB  =  CF=  FE.     (Horn,  sides  of  ^  A.) 

.-.  the  chord  ^5=  the  chord  CE.  (Ax.  2.) 

Q.E.D. 

IX.  3.  Li  the  same  circle,  or  in  equal  circles,  equal 
angles  at  the  center  subtend  equal  arcs  on  the  circum- 
ference, and  conversely. 

Hyp.     If,    in    the 
equal  (D  K  and  A^ 

Cone. :  then  arc  AB  =  arc  CE. 

Dem.  Place  Z  K  in  coincident  superposition  with  its  equal 
Z  Ki,  A  falling  on  (7;  then  B  must  fall  on  E.        '       '  (Ax.  7.) 

KA=KiC  and  KB=KiE,  being  radii  of  circles  equal  by  hyp. 

.-.  arc  AB  will  coincide  with  arc  CE.    (The  CD  are  =  by  hyp.) 

Q.E.D. 
,  Converse.     Proof  of  the  converse  is  left  to  the  pupil. 

IX.  3  a.  In  the  same  circle,  or  in  equal  circles,  equal 
chords  subtend  equal  arcs,  and  conversely. 

Ex.  1.  In  any  circle  the  greater  of  two  arcs  is  subtended  by  the  greater 
chord,  and  conversely.     (Use  IX.  3,  and  Ex.  9,  p.  65.) 

Ex.  2.  In  any  circle,  of  two  unequal  chords  the  one  nearer  the  center 
is  the  greater,  and  conversely. 

Let  AB  and  CE  be  the  chords,  KG,  the  distance  of  AB  from  the  cen- 
ter iT,  beina;  greater  than  KF,  the  distance  of  CE  from  K. 

On  KG  lay  off  KM  =  KF,  and  draw  through  M,  HL  i.  KM.  HL  =  CE 
(iX.  2).  But  arc  HABL  >  arc  AB  by  tlie  sum  of  arcs  HA  and  B.L. 
Hence,  chord  HL  >  chord  AB  by  Ex.  1. 

Ex.  3/'  If  two  circles  are  concentric,  tangents  to  the  first  circle  that  are 
chords  of  the  second  are  equal. 

Ex.  4.  If  a  radius  can  bedrawn  bisecting  the  angle  between  two  inter- 
secting chords,  the  chords  are  equal. 


82  THE  ELEMENTS  OF  GEOMETRY 

Tangency 
IX.  4.  A  radius  to  a  point  of  tangency  is  perpendicu' 


E     iV 


lar  to  the  tangent 

Hyp.     If  TN  is 

a  tangent,  and  KT 
is  a  radius  to  the 
point  of  tangency, 


Cone. :  then  TKl.  TN. 

Dem.     Let  KE  be  the  join  of  A"  and  any  point  of  T^  except  T. 
Then  E  must  be  without  the  circle.  (Def.  of  tangent.) 

.%  KT<KE.  (Def.  of  O.) 

That  is,  KT  is  shorter  than  any  other  line  from  K  to  TN. 

.-.  KTJL  TN.  (IV.  4  a,  Sch.) 

Q.E.D. 

IX.  4:  a.  A  line  perpendicular  to  a  radius  at  the  outer 
extremity  of  the  radius  is  the  tangent  to  the  circle  at 
that  point. 

Hyp.  If  TK  is  a 
radius,  and  if  TN 
LTK&tT, 

Cone. :  then        TN  is  tangent  to  O  K  at  T. 
Dem.     If  TN  is  not  tangent,  draw  TG,  that  is. 
Then  TK±  TG. 

But  TN±  TIC 

.'.  TN  must  coincide  with  T0» 
But  TG  is  tangent  to  O  A"  by  construction* 
.*.  TN  is  tangent 


ax.  4.) 

(Hyp.) 
(Ax.  7.) 

Q.B.D 


IX.     THE  CIRCLE   AND   ITS   RELATED  LINES  88 

IX.  4:b.  A  perpendicular  to  a  tangent  at  the  point 
of  tangency  passes  through  the  center  of  the  circle. 

Hyp.  If  TNi^ 
tangent  to  O  K, 
and  if  TKA.  TN 
at  the  point  of 
tangency, 

Cone. :  then.  T.S' passes  through  the  center  of  the  circle. 

Dem.  If  TK  does  not  pass  through  the  center,  draw  TF,  that 
does. 

Then  TF±TK  (IX.  4.) 

But  TK1.TN.  (Hyp.) 

.-.  TF  coincides  with  TK  (Ax.  7.) 

But  TF  was  drawn  through  the  center. 

.-.  T7r  passes  through  the  center.  Q.E.D. 

IX.  5.  Tangents  from  the  same  point  to  the  same 
circle  are  equal.  /^^T"'Nt>^ 

Hyp.     If  PG  i      Kv j— -™^P 

and  Pr  are  tan-  V  \     J^^^^ 

gents  to  O  K,  ^^ — *q 

Cone:  then  PG  =  PT. 

Dem.  Draw  radii  to  the  points  of  tangency  T  and  G ;  also 
draw  PK.  py  j_  j^y .  _pQ  _l  j^q  (ix.  4.) 

rt.  A  PKT  ^  rt.  A  PGK. 
[Two  right  A  are  ^  if  hypotenuse  and  leg,  etc.]  (V.  4.) 

.-.  PG  =  PT.         (Horn,  sides  of  ^  A.) 

Q.E.D. 

IX.  5  a.  The  join  of  the  center  and  the  iiitersection 
of  tioo  tangents  is  the  bisector  of  the  angle  made  hy  the 
tangents,  and  of  the  angle  made  hy  the  radii  to  the 
points  of  tangency. 


84 


THE  ELEMENTS  OF  GEOMETRY 


IX.  6.  If  two  circles  intersect,  the  line  of  centers  is 
the  mid-perpendicular  of  the  common  chord. 


Hyp.    If  0  iT 

intersects  O  Ki 
in  the  points  A 
and  Bj 


Cone. :  then  KK^  is  mid-_L  to  the  common  chord  AB. 

Dem.    Draw  the  radii  KA,  KB,  and  K^A  and  K^B. 

K  is  equidistant  from  A  and  B.      {KA  and  KB  being  radii.) 

Ki  is  equidistant  from  A  and  B,      {K^A  and  K^B  being  radii.) 

.'.  KK^  is  a  mid-perpendicular  to  AB. 

[Two  points  equidistant  from  the  ends  of  a  line  fix  the  mid- 
perpendicular  to  the  line.]  (VIII.  2,  Sch.  2.) 

Q.E.D. 

IX.  7.  i/*  ^ic;o  circles  are  tangent,  their  centers  and 
the  point  of  tangency  are  in  the  same  straight  line. 


Hyp.     If    Oir 

is  tangent  to  O  Ki 
at  T, 


Cone. :  then  K,  T,  and  K^  are  in  the  same  straight  line. 

Dem.     Draw  NG^  a  common  tangent  through  the  common 
point  T.  (Def.  of  tangent  ©.) 

KTJlNG. 

[A  radius  to  point  of  tangency  is  ±  to  tangent.]         (IX.  4.) 

KiT±NG.       (For  the  same  reason.) 

/.  -KTand  KiT  Sire  in  the  same  straight  line.     (Ax.  7.) 

Q.E.D 


IX.    THE   CIRCLE  AND  ITS  RELATED  LINES 


85 


IX.  8.  If  two  circles  are  tangent,  the  distance  hetiveen 
their  centers  is 

(a)  the  Slim  of  the  radii,  if  the  tangency  is  external ; 

(h)  the  difference  of  the  radii,  if  the  tangency  is  in- 
ternal. 


Fig.  2, 


Hyp.  If  ^1  and  7^2 
are  tangent  circles 
at  the  point  T,  and 
their  radii  are  r^ 
and  ^2, 


Cone:  then  (Fig.  1)  7^1/^2 =ri+r2  and  (Pig.  2)  KJC2=ri—r2. 

Dem.     Draw  the  common  tangent  LM  through  T. 

(Def.  of  tangent  (D.) 
Then  Ki,  K^,  and  T  are  in  the  same  straight  line.        (IX.  6.) 

.-.  (a)  K,T-\-  TK^^KJ^:^,  (Ax.  4.) 

and  (6)  K^  T-  TK^  =  K^K^ ;  (Ax.  4.) 

that  is,         (a)  K1K2  =  ^i  +  »'2  and  (&)  K1K2  =  i\  —  r^. 

Q.E.D. 

IX.  8  a.   Conversely.    If    the    distance    hetiveen    the 
centers  of  tivo  circles  is  equal  to 

(a)  the  sum  of  the  radii,  the  circles  are  tangeiit  to 
each  other  externally  ; 

(b)  the  difference  of  the  radii,  one  circle  is  tangent  to 
the  other  internally. 


Ex.  5.    Show  that  if  the  distance  between  the  centers  of  two  circles  is 

(a)  greater  than  the  sum  of  the  radii,  each  circle  is  without  the  other ; 

(6)  less  than  the  sum  of  the  radii,  but  greater  than  their  difference,  the 
circles  intersect  each  other  ; 

(c)  less  than  the  difference  of  the  radii,  one  circle  i&  wholly  within  the 
other. 


86  THE  ELEMENTS   OF   GEOMETRY 

IX.     SUMMARY   OF   PROPOSITIONS   IN   THE   GROUP 
ON   THE  CIRCLE  AND   ITS   RELATED   LINES 

1.  A  radius  perpendicular  to  a  chord  bisects  the  chord 
and  its  subtended  arc,  and  conversely, 

a   A  radius  perpendicular  to  a  chord  is  mid-per- 
pendicular to  every  chord  parallel  to  the  first 

2.  In  the  same  circle,  or  in  equal  circles,  equal  chords 
are  equidistant  from  the  center,  and  conversely. 

3.  Li  the  same  circle,  or  in  equal  circles,  equal  angles 
at  the  center  subtend  equal  arcs  on  the  circumference, 
and  conversely. 

a  In  the  same  circle,  or  in  equal  circles,  equal 
chords  subtend  equal  arcs,  and  conversely. 

4.  A  radius  to  a  point  of  tangency  is  perpendicular 
to  the  tangent. 

a    A  line  perpendicular  to  a  radius  at  the  outer 

extremity  of  the  radius  is  the  tangent  to  the 

circle  at  that  point, 
h   A  perpendicular  to  a  tangent  at  the  point  of 

tangency  passes  through  the  center  of  the 

circle. 

5.  Tangents  from  the  same  point  to  the  same  circle 
are  equal. 

a  The  join  of  the  center  and  the  intersection  of 
two  tangents  is  the  bisector  of  the  angle  made 
by  the  tangents,  and  of  the  angle  made  by 
the  radii  to  the  points  of  tangency. 


IX.     THE    CIRCLE   AND   ITS   RELATED   LINES  87 

6.  If  huo  circles  intersect,  the  line  of  centers  is  the 
mid-perpendicular  of  the  common  chord. 

7.  If  two  circles  are  tangent,  their  cejiters  and  the 
point  of  tajigency  are  in  the  same  straight  line. 

8.  If  two  circles  are  tangent,  the  distance  hetiveen 
their  centers  is 

(a)  the  sum  of  the  radii,  if  the  tangency  is  external ; 
(5)  the  difference  of  the  radii,  if  the  tangency  is 
iiiternal, 

8  a.  Conversely.  If  the  distance  hetween  the  centers 
of  two  circles  is  equal  to 

(a)  the  sum  of  the  radii,  the  circles  are  tangent  to 

each  other  externally  ; 

(b)  the  difference  of  the  radii,  one  circle  is  tangent 

to  the  other  internally. 

The  Isosceles  Triangle  as  Part  op  the  Sector  of  a 
Circle 

Note. — In  an  isosceles  triangle,  the  altitude  to  the  base  is  identical 
with  the  median  to  the  base.  (IV.  1  a,  Sch.) 

Note.  — Observe  that  if  the  vertex  of  the  vertex  angle  of  an  isosceles 
triangle  be  taken  as  a  center  and  a  circle  be  described  with  either  leg  as 
a  radius,  the  legs  of  the  triangle  are  radii  of  the  circle ;  the  base  of  the 
triangle  is  a  chord,  and  the  altitude  to  the  base,  the  median,  and  the 
bisector  of  the  vertex  angle  (which  we  have  seen  in  (IV.  3)  to  be  the  same 
line)  are  a  part  of  the  radius  perpendicular  to  the  chord. 


88 


THE  ELEMENTS  OF  GEOMETRY 


PROBLEMS 


Prob.  I.    To  construct  a  tangent  through  a  given  point 
to  a  given  circle. 


Given.     A  circle  Ky  and  a  point  P. 

Required.     To  construct  a  tangent  through  P  to  the  circle  K. 
Case  I.     When  P,  the  given  point,  is  on  the  circle. 
What  is  the  angle  formed  by  a  tangent  and  a  radius  drawn 
to  the  point  of  tangency  ?    What  is  the  construction  required  ? 
Case  II.     When  P,  the  given  point,  is  without  the  circle. 

Const.    Join  P  and  K,  and  on  PK  as  a  diameter  describe  a 
circle  intersecting  the  given  circle  in  T  and  G. 
Then  PT  and  PG  are  the  required  tangents. 

Proof.     Let  the  pupil  supply  the  proof. 
-     Why  is  Z  PTK  a  right  angle  ? 

Prob.  II.     To  construct  a  common  exterior  tangent  to 
two  given  circles  ivhose  radii  are  r  a7id  r. 

Given.     Two  circles,  K  and  /fi,  

whose  radii  =  r  and  r'. 

Required.     To  construct  a  com- 
mon exterior  tangent. 

Const.     With  if  as  a  center  and  a  radius  =r  —  r',  draw  the 
inner  concentric  circle. 

From  Ki  draw  a  tangent  to  this  circle.  (Prob.  I.) 


Q.E.F. 

(VII.  4  rt.) 


IX.    THE   CIRCLE   AND   ITS   RELATED   LINES  89 

Draw  KT  to  the  point  of  tangency,  T,  and  produce  it  to  meet 

the  outer  circle  in  N. 

Through  iV^draw  NL  II  TK,.  (Prob.  IV.,  p.  50.) 

Through  K^  draw  a  line  II  /0»''and  meeting  NL,  say  at  6r. 

KG  is  the  reqitired  common  exterior  tangent. 

Q.E.F. 

Proof.     The  4-side  T—  G  is  a  parallelogram.     (Def.  of  a  O.) 

.-.  K^G  =  TN. 

[Opposite  sides  of  a  parallelogram  are  equal.]         (YI.  1  a.) 
But  TN=r'  by  construction. 

.-.  K,G  =  r'  (Ax.  1.) 

But  Z  T  is  a  right  angle.  (IX.  4.) 

.-.  all  the  A  of  the  4-side  T—  G  are  right  angles.    (VI.  1'  a'.) 
.*.  iVGi^  is  a  common  exterior  tangent  to© /f  and  ^1.  (IX.  4  a.) 

Q.E.D. 

Note.  —  Another  common  tangent  may  be  found,  crossing  KKi  between 
the  circles,  and  therefore  called  an  interior  tangent. 

In  this  case  the  first  auxiliary  circle  has  the  radius  =  r  +  r'  instead  of 
r  —  r'.  Let  the  student  give  the  construction  in  full.  Show  that  four  com- 
mon tangents  are  possible.  When  may  three  only  be  drawn  ?  When 
two  ?     When  one  ?     When  none  ? 

Ex.  6.    (a)  Draw  a  chord  equal  and  parallel  to  a  given  chord. 
(6)   Draw  a  chord  equal  and  perpendicular  to  a  given  chord. 

Ex.  7.  If,  in  a  circle,  two  equal  chords  are  drawn,  and 
a  radius  is  drawn  to  the  end  of  each  chord,  the  angles 
between  the  radii  and  the  chords  are  equal  to  each 
other.  c 

Ex.  8.  In  the  same  or  equal  circles  the  greater  of  two  minor  arcs  is 
subtended  by  the  greater  chord. 

Ex.  9.   If  chord  ^i?'=  chord  BC,  then  arc  AB  = 
arc  CF. 


90 


THE  ELEMENTS  OF  GEOMETRY 


Ex.  10.  Show  that  two  chords  that  are  not  diameters  cannot  bisect  each 
other. 

Ex.  11.  Prove  by  means  of  IX.  3  a  that  two  triangles  are  congruent  if 
tliree  sides  of  the  one  equal  three  sides  of  the  other. 

Ex.  12.  If  an  inscribed  polygon  is  equiangular,  it  is  not  necessarily 
equilateral.  (y 


Ex.  13.   If,  in  Fig.  1,  the  secant  is  drawn  so  that  ^ 
AB  =  BKy  show  that  Z  CKF  =  3  Z  J. 


Ex.  14.  Two  parallel  chords  intercept  equal  arcs 
on  the  circumference.     (Fig.  2.) 

Ex.  15.  A  chord  and  a  parallel  tangent  inter- 
cept equal  arcs  on  the  circumference. 

Ex.  16.    Draw : 

(1)  Two  common  exterior  tangents  to  two  circles. 

(2)  Two  common  interior  tangents  to  two  circles. 
Ex.  17.   Prove  that  the  above  exterior  tangents  are  equal. 

Ex.  18.    In  Fig.  3,  O  Xis  tangent  to  O  i^  at  T,  and  is  also  tangent  to  LP 
at  P. 

Why  are  K^  X,  and  T  in  the  same  straight 
line? 

Ex.  19.  If  O  X  is  tangent  to  LF  at  P, 
why  is  Z  LPA  a  right  angle  ? 

Ex.  20.  If  PA  =  KT,  why  is  A  KAX 
.isosceles  ? 

Ex.  21.  How,  then,  if  PA  and  KA  are 
given  in  position  and  length,  may  the  points 
X  and  T  be  determined  ? 

Ex.  22.   Problem :  Given  the  line  LF,  the 
point  P  in  LF,  and  the  O  K,  construct  a  circle  that  shall  be  tangent  to 
LF  a,t  P  and  also  tangent  to  O  K. 

Ex.  23.  Similarly,  by  laying  off  PA' 
(=  PA)  below  LP,  find  the  center  X  of  a 
second  circle  that  shall  also  be  tangent  to 
LF  at  P  and  likewise  tangent  to  O  K. 

Ex.  24.  Show  that  the  hypotenuse  of  a 
rt.  A  equals  the  sum  of  the  two  remaining 
sides,  minus  twice  the  radius  of  the  in- 
scribed O. 


X.     GROUP   ON   CONCURRENT   LINES   OF 
A   TRIANGLE 

DEFINITION 

Lines  are  Concurrent  when  they  intersect  at  the  same  point. 
PROPOSITIONS 

X.  1.   The  bisectors  of  the  interior  angles  of  a  triangle 
concur. 


Hyp.     If 

ABC  is    a 
triangle, 


Cone. :  then  the  bisectors  oi  Z.  A,  Z.  B,  and  Z.  C  concur. 
Dem.     The  bisector  of  Z  ^. either  meets,  or  is  parallel  to,  the 
bisector  of  Z  C.  /  a       /  n 

If  the  bisectors  are  II,  the  ^  +  ^  =  2  rt.  A  (II.  2.) 

.'.  ZA  +  ZC  =  4:  rt.  A.  (Ax.  3.) 

But  this  conclusion  is  impossible,  (III.  1.) 

.'.  the  bisector  of  Z  A  must  meet  the  bisector  of  Z  C. 
Let  the  point  of  intersection  be  Ki. 

Ki,  in  the  bisector  of  Z  C,  is  equidistant  from  a  and  b,  and 
Ki,  in  the  bisector  of  Z  A,  is  equidistant  from  b  and  c. 

[Every  point  in  the  bisector  of  an  angle,  etc.]  (VIII.  3.) 

.*.  Ki  is  equidistant  from  a  and  c.  (Ax.  1.) 

.-.  Ki  is  in  the  bisector  of  Z  B.  (VIII.  4,  Sch.) 

.'.  the  bisectors  of  ZA,ZB,  and  Z  C  concur. 

Q.E.D. 
This  point  of  concurrence  is  called  the  In-center. 

91 


92 


THE  ELEMENTS  OF  GEOMETRY 


X.  1  a.  If  the  in-center  he  taken  as  a  center,  and  the 
distance  to  any  side  as  a  radius,  a  circle  may  he  drawn 
tange7it  to  the  sides  of  the  triangle. 

X.  2.  .The  hisectors  of  one  interior  angle  of  a  triangle 
and  the  two  exterior  angles  non-adjacent  to  it  concur, 

Cy 


Hyp.    If 

ABC  is   a 

triangle, 


Cone. :  then  the  bisectors  of  ext.  Z  A,  ext.  A  By  and  int.  Z  C 
concur. 

Dem.     The  bisector  of  Z  (7  either  meets,  or  is  parallel  to,  the 
bisector  of  ext.  Z  A, 

'  If  parallel,  then  ^^LAA  =  ^. 

[If  two  parallels  be  crossed  by  a  third  line,  etc.]         (II.  1.) 

.-.  ext.  Z^  =  Za  (Ax.  3.) 

But  this  conclusion  is  impossible, 

[The  exterior  angle  of  a  A  is  greater,  etc.]         (III.  2,  Sch.) 
.-.  the  bisectors  of  ext.  A  A  and  of  Z  C  must  meet,  say  at  K^, 
K^,  in  the  bisector  of  Z  C,  is  equidistant  from  a  and  b  (pro^ 
duced). 

[Every  point  in  the  bisector  of  an  angle,  etc.]         (VIII.  3.) 
K^,  in  the  bisector  of  ext.  A  A,  is  equidistant  from  c  and  b. 

(Same  reason.) 
. ,  Kfis  equidistant  from  c  and  a.  (Ax.  1.) 

.*.  K^  must  lie  in  the  bisector  of  ext.  Z  B. 


X.     CONCURRENT  LINES   OF  A  TRIANGLE  93 

[The  bisector  of  an  angle  is  the  locus,  etc.]      (VIII.  4,  Sch.) 
.*.  the  bisectors  of  ext.  Z.A,  ext.  Z  B,  and  int.  Z  C  concur. 

Q.E.D. 

The  point  of  concurrence  is  called  an  Ex-center  of  the  tri- 
angle. There  are  two  other  ex-centers :  viz.,  K^.  on  the  bisector 
of  Z  A,  and  K^..  on  the  bisector  oi  Z.B. 

Sch.  To  three  non-concurrent  lines,  three  tangent  circles, 
(besides  the  inscribed  circle  already  indicated)  may  be  drawn, 
with  K^,  K^',  Ke"  as  centers,  and  the  distances  from  these  points 
to  the  corresponding  lines  as  respective  radii. 

Circles  tangent  to  one  side  of  a  triangle  and  to  the  two  other 
sides  produced,  are  called  Escribed  Circles. 


Ex.  1.   In  A  ABC,  I  is  the  center  of  the  in- 

scribed  circle.     Show  that  Z  /  =  rt.  Z-\ ^• 

2 


Ex.  2.   In  A  ABC,  E  is  the  center  of  the  escribed 

circle.     Show  that  Z  ^  =  rt.  Z  -  — • 

2 

Prove  that  ZE  oi  Fig.  2  is  the  supplement  of 
Z/of  Fig.  1. 


Find  the  locus  of  the  center  of  a  circle  that  satisfies  the  following 
conditions  : 

Ex.  3.   That  passes  through  Q,  and  has  a  radius  a. 

Ex.  4.    That  touches  a  line  AL,  and  has  a  radius  a. 

Ex.  5.   That  touches  a  circumference,  and  has  a  radius  a. 

Ex.  6.    That  passes  through  two  given  points. 

Ex.  7.    That  is  tangent  to  two  intersecting  lines. 

Ex.  8.   That  is  tangent  to  two  parallel  lines. 

Ex.  9.   That  is  tangent  to  two  equal  circumferences. 

Ex.  IQ.   That  is  tangent  to  a  line  at  a  given  point. 

Ex.  11.    That  is  tangent  to  a  circumference  at  a  given  point. 


94  THE  ELEMENTS  OF  GEOMETRY 

X.  3.   The  mid-perpendiculars  to  the  three  sides  of  a 
trimigle  concur. 

Hyp.  If 
ABC  is  a 
triangle, 

Cone. :  then  the  mid-perpendiculars  to  a,  6,  and  c  concur. 

Dem.     The  mid-perpendicular  to  a  either  intersects,  or  is 
parallel  to,  the  mid-perpendicular  to  h. 

If  they  be  parallel,  AG  and  CB  would  have, 

(1)  the  same  direction,  •.•  both  would  be  perpendicular  to 

these  parallels ;  and 

(2)  the  point  G  in  common. 

.'.  AG  and  GB  would  lie  in  the  same  straight  line.     (Ax.  7.) 
But  this  conclusion  is  contrary  to  the  hypothesis. that  ABG 
is  a  triangle. 

.-.  the  mid-perpendicular  to  a  must  intersect  the  mid-perpen- 
dicular to  h  in  some  point,  say  K,.. 

K^,  in  the  mid-perpendicular  to  a,  is  equidistant  from  (7 and  B. 

[Every  point  in  the  mid-perpendicular,  etc.]  (VIII.  1.) 

ifg,  in   the   mid-perpendicular  to  6,  is  equidistant  from  G 

and  A.  (Same  reason.) 

.'.  -H^c  is  equidistant  from  A  and  B.  (Ax.  1.) 

.•.  Kc  is  in  the  mid-perpendicular  to  c. 

[The  mid-±  to  a  line-segment,  etc.]  (VIII.  2,  Sch.) 

.-.  the  mid-perpendiculars  concur. 

Q.E.D. 

The  point  of  concurrence  is  called  the  Circumcenter. 

Sch.     Through  three  non-collinear  points  an  unique  circle 

may  be  drawn. 


X,  CONCURRENT  LINES  OF  A  TRIANGLE      95 

X.  4.   The  altitudes  of  a  triangle  concur. 


Hyp.     If 

ABC   is   a 
triangle, 


^7""- 


^1 
Cone. :  then  the  altitudes  AL,  BT,  and  C/ concur. 

Dem.  Through  C,  A,  and  B  draw  parallels  to  c,  a,  and  6, 
respectively. 

Produce  these  parallels  to  intersect  in  Cj,  A^,  and  By. 

The  4-sides  ABCB^  and  ABA^C  are  ZI7.  (Def.  of  O.) 

.-.  ^iC=  ^S;    CA  =  AB.  (VI.  1  a.) 

.'.  B,C=CAi.  (Ax.  1.) 

Again,  (7/±  BiA^.     (Def.  of  lis,  direct  inference.) 

.*.  CI  is  the  mid-perpendicular  to  ByAy, 

Similarly,  BT  is  mid-perpendicular  to  A^'ij  -^-^j  mid-per- 
pendicular to  BiCi. 

.'.  the  altitudes  of  the  original  triangle  are  mid-perpendicu- 
lars to  the  sides  of  the  larger  triangle. 

But  these  mid-perpendiculars  concur.  (X.  3.) 

.*.  the  altitudes  of  the  original  triangle  concur. 

Q.E.D. 
The  point  of  concurrence  is  called  the  Orthocenter. 

Construct  a  circle  that  satisfies  the  following  conditions : 

t 

Ex.  12.    That  has  a  given  radius  a,  and  passes  through  two  given  points. 

Ex.  13.  That  has  a  given  radius  a,  and  is  tangent  to  two  given  inter- 
secting lines  AL  and  BM. 

Ex.  14.  That  has  a  given  radius  a,  and  is  tangent  to  two  equal  cir- 
cumferences. 

Ex.  15.  That  has  a  given  radius  a,  and  is  tangent  to  a  given  line  at  a 
given  point. 


96  THE  ELEMENTS  OF  GEOMETRY 

X.  6.   The  medians  of  a  triangle  concur. 


Hyp.     If 

ABC  is   a 
triangle, 

Cone. :  then  the  medians  AH,  BL,  and  CF  concur. 

Dem.     CF  and  BL  intersect,  or  are  parallel. 
If  CF  II  BL,  all  points  in  CF  must  lie  on  the  same  side  of  BL. 
But,  as  a  consequence  of  the  definition  of  a  median,  BL  must 
lie  between  c  and  a. 

,'.  C  and  F  must  lie  on  opposite  sides  of  BL. 

.'.  CFand  BL  must  intersect,  say  at  Kg. 

Draw  AKg ;  draw  LJ  and  FO II  ^^, ;  also  LF,  and  OJ". 

Then  i  J  and  i?^0  each  II  AK„  and  =  ^. 

[If  in  a  A  a  parallel  to  the  base,  etc.]  (VII.  3  c.) 

.'.  the  4-side  LJOF  is  a  parallelogram. 

[A  4-side  is  a  parallelogram  if  it  has  one  set,  etc.]      (VI.  2.) 

.-.  KgF=KgJ. 

[The  diagonals  of  a  parallelogram  mutually  bisect.]    (VI.  3.) 

Now  JKg^JC.  (Const.) 

.-.  KgF  =  ^.     Similarly,  K,L  =  ^ . 

.*.  as  any  two  medians  cut  off  the  same  third  of  the  third 

median,  the  three  medians  must  concur. 

Q.E.D 

This  point  of  concurrence  is  called  the  Center  of  Gravity,  oi 

Centroid,  of  the  triangle. 

Ex.  16.   Construct  a  circle  that  has  a  given  radius  a,  and  is  tangent  to 
a  given  circumference  at  a  given  point. 


X.  CONCURRENT  LINES  OF  A  TRIANGLE      97 

Ex.  17.  That  has  a  given  radius  a,  passes  through  a  given  point  Q, 
and  is  tangent  to  a  given  line  AL. 

Ex.  18.  That  has  a  given  radius  a,  passes  through  a  given  point  Q, 
and  is  tangent  to  a  given  circumference. 

Ex.  19.  That  has  a  given  radius  a,  is  tangent  to  a  given  line  AL,  and 
is  also  tangent  to  a  given  circumference. 


X.     SUMMARY   OF  PROPOSITIONS   IK  THE   GROUP 
ON    CONCURRENT  LINES   OF   A  TFIANGLE 

1.  The  bisectors  of  the  interior  angles  of  a  triangle 
concur, 

a  If  the  in-center  he  taken  as  a  center,  and  the 
distance  to  any  07ie  side  as  a  radius,  a  circle 
may  he  drawn  tangent  to  the  sides  of  the 
triangle. 

2.  The  hisectors  of  one  interior  angle  of  a  triangle 
and  the  two  exterior  angles  non-adjacent  to  it  concur. 

ScH.  To  three  non-concurrent  lines  three  tangent  circles 
may  be  drawn,  with  K^,  K^.,  K^..  as  centers  and  the  distances 
from  these  points  to  the  corresponding  lines  as  respective  radii. 

3.  The  mid-perpendiculars  to  the  three  sides  of  a  tri- 
angle concur, 

4.  The  altitudes  of  a  triangle  concur. 

ScH.  Through  three  non-collinear  points  an  unique  circle 
may  be  drawn. 

5.  The  medians  of  a  triangle  concur. 


98 


THE  ELEMENTS  OF  GEOMETRY 


Fig.  1. 


X.     CONCURRENT  LINES   OF  A  TRIANGLE  99 

SUMMARY  OF  TRIANGULAR  RELATIONS 
Important  Properties  of  the  Angles  of  a  Triangle  (Fig.  1) 

1.  ZA  +  ZB-{- ZC  =  2Tt.  A.  (in.  1.) 

2.  ZAis  the  supplement  ot  Z  B  -{-  Z  0.  (III.  1  a. ) 

3.  If  Z  C  is  a  right  angle,  Z  B  is  the  complement  of  Z  A.     (III.  16.) 

4.  liZA  =  ZB,  a  =  b.  (IV.  2.) 

5.  ItZA  =  ZB=:ZC,a  =  b  =  c. 

6.  IiZA>ZB,  a>b.  (IV.  5.) 

7.  Ext.  ZA  =  ZB-\-ZC.  (in.  2.) 

8.  Ext.  ZA>  ZB  or  ZC.  (IIL  2,  Sch.) 

9.  It  ZA  =  ZB,  ext.  ZC  =  2ZAor2ZB. 

10.   The  shape  (not  size)  of  a  triangle  is  given  by  any  two  of  its  inde- 
pendent angles. 

Important  Properties  of  the  Lines  op  a  Triangle  (Fig.  2) 

1.  a  +  b>c.  (VIL  1.) 

2.  lta  =  b,ZA=ZB.  (IV.  1.) 
3.lfa  =  b  =  c,ZA  =  ZB  =  ZC. 

4.  The  bisector  of  Z  J.  is  perpendicular  to  the  bisector  of  ext.  Z  A. 

5.  The  bisectors  of  ZA,   Z  B,   and  Z  C  concur  at  the  in-center. 

(X.L) 

6.  The  bisectors  of  ext.  Z  A,  ext.  Z  B,  and  of  Z  C  concur  at  an 
ex-center.  (X.  2.) 

7.  The  mid-Js  to  a-,  &,  and  c  concur  at  the  circumcenter.  (X.  3.) 

8.  The  altitudes  to  a,  b,  and  c  concur  at  the  orthocenter.  (X.  4.) 

9.  The  medians  to  a,  6,  and  e  concur  at  the  centroid.  (X.  5. ) 
10.    The  shape  and  size  of  a  triangle  are  determined  by  any  three 

independent  parts. 

Important  Properties  of  the  Lines  op  a  Right  Triangle  (Fig.  3) 

1.  The  median  to  the  hypotenuse  equals  J  the  hypotenuse.     (VII.  4.) 

2.  If  Z  ^  =  I  rt.  Z,  the  median  to  the  hypotenuse  equals  b. 

3.  The  altitude  to  a  coincides  with  &,  and  vice  versa. 

4.  The  hypotenuse  is  the  diameter  of  the  circumcircle,  (VII.  4.) 

5.  {a  -{-  b)—  c  is  the  diameter  of  the  inscribed  circle. 

(Ex.  24,  p.  90.) 

6.  The  hypotenuse,  c,  >  a  or  b.  (IV.  4  a.) 


XI.     GROUP   ON  MEASUREMENT 

DEFINITIONS 

Measurement  is  (a)  Direct,  or  (b)  Indirect.* 

The  Direct  Measurement  of  a  magnitude  is  the  process  of 
finding  how  many  times  it  contains  another  magnitude  of  the 
same  kind,  which  is  called  the  unit  of  measure. 

E.g.  length  may  be  measured  in  feet,  miles,  meters,  kilo- 
meters, etc. ;  area  in  acres,  square  miles,  hectares,  etc.,  and 
weight  in  kilograms,  pounds,  tons,  etc. 

Any  line  may  be  assumed  as  a  Unit  of  Length. 

The  Indirect  Measurement  of  a  quantity  is  the  process  of  de- 
termining its  size  by  comparing  it  with  some  other  quantity,  the 
changes  in  size  of  which  correspond  to  changes  in  size  of  the 
first  magnitude. 

E.g.  the  pressure  of  steam  is  measured  by  the  changes  in 
position  of  a  hand  on  a  dial  plate. 

The  height  of  a  mountain  is  measured  by  the  motion  of  the 
index  on  an  aneroid  barometer. 

The  strength  of  an  electric  current  is  measured  by  the  tem- 
perature to  which  it  raises  a  wire  of  known  dimensions. 

The  pitch  of  an  organ  pipe  is  measured  by  the  length  of  the 
pipe. 

The  amount  of  acid  in  a  solution  is  measured  by  the  intensity 
of  the  color  it  produces  in  a  piece  of  litmus^  paper. 

Angular  Measure.  Among  geometrical  magnitudes  an  angle 
is  often  measured  by  its  intercepted  arc,  or  by  the  quotient 
of  the  intercepted  arc  divided  by  the   radius  of  the  circle 

'      1  Whether  measurement  is  direct  or  indirect,  the  object  attained  is  the 
same.    The  measure  of  a  magnitude  is  a  ratio ;  therefore,  always  abstract. 

100 


XI.     MEASUREMENT  101 

whose  center  is  the  vertex  of  the  angle.     The-^tjber;!?  cal!(?d; 

Radial   Measure.  ,>>.,.,  ^  ^  >',  ^ 

Ratio  and  Proportion  ;  ' '  - ,  -  • ''  '  '' 

The  Geometric  Ratio  of  one  magnitude  to  another  is  the 
quotient  obtained  by  dividing  the  first  by  the  second. 

Thus,  the  ratio  of  a  to  6  is  -  =  a  h-  &  =  a  :  6. 

b 

The  Antecedent  of  a  ratio  is  the  first,  or  dividend  magnitude. 

The  Consequent  of  a  ratio  is  the  second,  or  divisor  magni- 
tude. 

The  usual  Sign  of  ratio  is  :,  although  any  method  of  indi- 
cating division  may  be  used. 

Both  terms  of  any  ratio  may  be  multiplied  or  divided  by  the 
same  quantity,  without  affecting  the  value  of  the  ratio. 

Commensurable  Ratios 

When  the  terms  of  a  ratio  can  each  be  expressed  as  a  multi- 
ple of  a  common  unit,  the  terms  are  said  to  be  commensurable 
with  each  other,  and  the  ratio  is  said  to  Commensurable. 

In  this  case  the  ratio  can  always  be  expressed  as  a  numerical 
fraction,  both  of  whose  terms  are  whole  numbers. 

Problem  I.   To  express  the  ratio  of  tivo  line-segments. 

Given.     AB  and  CE.  , L_ 

Z  GB 

Required.     The  ratio 

AB.CE.  G Hi 

Apply  CE  to  AB  as  often  as  possible,  say  twice,  with  a 
remainder  GB,  so  that 

AB  =  2CE-\-GB. 

Apply  the  remainder  GB  to  CE  as  often  as  possible,  say  four 
times,  with  a  remainder  FE,  so  that 

CE=:4.GB  +  FE, 
and  AB  =  SGB  +  2FE+GB 

=:9GB  +  2FE. 


102         THE  ELEMENTS  OF  GEOMETRY 

fVpply  tht  'ffLSft  remainder  FE  to  OB  as  often  as  possible,  say 
fo\:^r .ti|iif)9,  withqut. remainder,  so  that 

''  '''•'>■'  ■''•  •''<'■''< '93=   ^FE, 

and  CE=16FE+    FE  =  17  FE, 

and  AB  =  36  FE  +  2  FE  =  SS  FE. 

Thus  the  given  lines  have  been  expressed  in  terms  of  the 
common  unit  FE,  and  their  ratio 

AB^3SFE^3S 
CE     17  FE     17* 

The  ratio  ^  =  ~ 

AB     38 

Apply  the  same  method  to  the  solution  of  the  following 
problems. 

Problem  II.   To  express  the  ratio  of  two  arcs  of  equal 
circles,  arc  AB  and  arc  CE. 

Problem  III.   To  express  the  ratio  of  two  angles  or 
sectors  of  equal  circles,  /.  AKB  and  Z  CK'E, 


^  Incohmensubablb  Ratios 

If,  on  the  other  hand,  the  terms  of  a  ratio  cannot  be  expressed 
as  multiples  of  the  same  unit,  the  terms  are  incommensurable 
with  each  other,  and  the  ratio  is  said  to  be  Incommensurable. 

In  this  case  no  one  of  the  remainder  line-segments,  arcs, 
angles,  etc.,  of  the  process  just  indicated  will  be  exactly  con- 
tained in  the  preceding  remainder  line-segment,  arc,  angle,  etc., 
no  matter  how  long  the  process  be  continued. 


XI.     MEASUREMENT  103 

Hence,  an  incommensurable  ratio  cannot  he  expressed  as  a 
fraction  the  terms  of  which  are  whole  numbers. 

Approximate  Expression  of  Incommensurable  Ratios 

Such  a  ratio  can  usually  be  expressed,  however,  in  some  form 
that  will  enable  us  to  state  the  value  of  the  ratio  correctly 
to  any  required  decimal  place,  or  to  any  required  degree  of 
accuracy. 

Hence,  all  the  ratios  with  which  we  shall  deal  may  be  ex- 
pressed, either  exactly  or  approximately,  as  fractions.  But  the 
necessity  for  thus  expressing  them  seldom  arises.  We  shall 
be  concerned  chiefly  with  the  relations  between  ratios,  and  one 
of  the  most  important  of  these  geometric  relations  is  that  of 
equality.     This  relation  gives  us  the  geometric  proportion. 

A  Geometric  Proportion  is  an  expression  of  equality  between 
two  or  more  geometric  ratios. 

Thus,  a:b  =  c  :m, 

or,  as  oftener  written,  a:b  ::  c:m, 

and  [jP:\JQ::h:h' 

are  geometric  proportions. 

If  more  than  two  ratios  are  compared,  the  proportion  is  said 
to  be  a  Continued  Proportion. 

Each  ratio  of  a  proportion  is  called  a  Couplet. 

The  Extremes  are  the  first  and  fourth  terms  of  a  proportion. 

The  Means  are  the  second  and  third  terms  of  a  proportion. 

It  a:b::c:  e,  e  is  said  to  be  a  Fourth  Proportional  to  a,  b, 
and  c.     Similarly,  6  is  a  fourth  proportional  to  a,  c,  and  e,  etc. 

li  a:b  ::b  :  c,  cis  said  to  be  a  Third  Proportional  to  a  and  b. 
Similarly,  a  is  a  third  proportional  to  b  and  c. 

li  a:  b::b:  c,  b  is  said  to  be  a  Mean  Proportional  between  a 
and  c. 

A  Transformation  of  a  proportion  is  a  change  in  the  propor- 
tion, either  in  the  order  of  the  terms  or  otherwise,  that  does 
not  destroy  the  equality  of  the  ratios. 


104         THE  ELEMENTS  OF  GEOMETRY 

A  Derived  Proportion  is  one  obtained  from  a  given  proportion 
by  transformation. 

Thus,  a^:b^i:c^i€^ 

is  derived  from  a  :b  ::c  :e 

by  cubing  the  terms  of  the  latter. 

PROPOSITIONS 

XL  1.  If  four  quantities  are  in  j^roportmiy  the  prod- 
uct of  the  means  equals  the  product  of  the  extremes, 
and  conversely. 

If  a:b:ic:e,  i.e.  if  -  =  ^, 

b     e 
then  ae  =  be. 

[If  both  members  of  the  given  equation  be  multiplied  by 
be,  the  results  will  be  equal.]  (Ax.  3.) 

Q.E.D. 

Conversely,  if  the  product  of  two  quantities  equals  the 
product  of  two  others,  either  set  of  factors  may  he  made 
the  extremes,  and  the  other  the  means,  of  a  proportion. 

Hyp.     If  ae  =  be, 

Cone. :  then  a:b::c:e,  i.e.  ^  =  -. 

b     e 

[If  both  members  of  the  given  equation  be  divided  by  be,  the 
results  will  be  equal.]  (Ax.  3.) 

Q.E.D. 

ScH.  1.  We  have  seen  that  in  the  original  proportion  the 
product  of  the  means  equals  the  product  of  the  extremes. 

The  test  of  the  coiTectness  of  every  derived  proportion  is 
that  when  the  product  of  its  extremes  is  placed  equal  to  the 
product  of  its  means,  the  resulting  equation  is  the  same  as  the 
equation  similarly  obtained  from  the  origincd  proportion,  or 
may  be  reduced  to  the  same. 


Hyp. 

If 

a  :  b  : :  c  :  e, 

Cone. : 

;  then  (1) 

b  :  a  : :  e  :  c. 

Dem. 

If 

«=:^  then  i  =  i. 
be             a     c 

b     e 

b__e_ 

"a     c 

XI.     MEASUREMENT  105 

Illustration.  —  If  a:b:  :  c:  e  be  the  original  proportion, 
then,  by  the  test,  a-{-b:b::c-\-e:e\s  a  correct  derived  pro- 
portion ;  for,  by  the  application  of  XI,  1  to  each,  we  get  in  each 
case  ae  =  be. 

(Let  the  student  make  the  application.) 

ScH.  2.  The  most  important  transformations  give  the  fol- 
lowing derived  proportions : 


(Ax.  3.) 


Q.E.D. 

(This  form  is  said  to  be  derived  from  the  given  proportion 
by  inversion.) 

Cone.  (2)  :  a  :  e  : :  b  :  e. 

Dem.    If  ^  =  ^,  then  ^x-  =  -x-.  (Ax.  3.) 

be  b     c     e     c 

-  —  -, 

'  c     e 

Q.E.D, 

(This  form  is  said  to  be  derived  from  the  given  proportion 
by  alternation.) 

Cone.  (3) :  a  +  b  :  a : :  c  4-  e :  c,  and  a-fb:b::c-He:e- 

Dem.     If  ^  =  ^,  then  ^  + 1  =  £  + 1.  (Ax.  2.) 

be  b  e 

a  +  &_c-f-e 

b     ~     e    ' 

Q.E.D. 

(This  form  is  said  to  be  derived  from  the  given  proportion 
by  composition.) 


106         THE  ELEMENTS  OF  GEOMETRY 

Cone.  (4)  :  a  —  b  :  a : :  c  —  e  :  c,  and  a  —  b  :  b  : :  c  —  e :  e. 

Dem.     If  ?  =  -,  then,  ?- 1=^-1.  (Ax.  2.) 

be  be 

"      b     "    e   ' 

Q.E.D. 

(This  form  is  derived  from  the  given  proportion  by  division.) 

Cone.  (5):  a  +  b:a  —  b::c  +  e:e  —  e. 

Dem.  Divide  the  last  equation  in  conclusion  (3)  by  the  last 
equation  in  conclusion  (4),  member  by  member. 

...  ?L±:^  =  ^±^.  (Ax.  3.) 

a—b     c—e 

Q.E.D. 

(This  form  is  said  to  be  derived  from  the  given  proportion 
by  composition  and  division.) 

XI.  2.  In  any  number  of  equal  ratios,  the  sum  of 
the  antecedents  is  to  the  sum  of  the  consequents  as  any 
antecedent  is  to  its  consequent. 

Hyp.     If  a:  6::  c:  e::/:gr::etc., 

Cone.:  then  a  +  c+/+ '•• :  &  +  «  +  g'-h  •••: :  a:  6::  c;  e  ::  etc. 

Let  ?  =  -  =  ^=^, 

beg 

whence  a  =  br\  c  =  er\  f=  gr,  etc., 

and  a +  C+/+ •••  =  5r  +  er  +  grr  +  •••. 

Then  a  +  c+f-i-  ...  =  r(6  + e +^+ ...). 

b  +  e-\-g-i b     e 

Q.E.D. 

XI.  3.  If  two  proportions  he  multiplied  together, 
term  by  term,  the  resulting  products  will  he  in  propor- 
tion. 

Let  the  student  supply  the  proof,  by  use  of  Ax.  3. 


XI.     MEASUREMENT  107 

a   If  any  number  of  propor^tions  he  multiplied 

together,  term  by  term,  the  resulting  products 

will  be  in  proportion. 
b   Like  powers  of  the  terms  of  a  proportion  are 

in  proportion. 
c   Like  roots  of  the  terms  of  a  proportion  are  in 

proportion. 

XI.  4.  If  the  terms  of  a  proportion  be  divided  suc- 
cessively by  the  terms  of  a  second,  the  residting  quotients 
will  be  in  proportion. 

Let  the  student  supply  the  proof,  using  Ax.  3. 


Ex.  1.  If  2/=  c,  what  is  the  ratio  of  c  to/? 
Ex.  2.  If  a  =  3  e,  what  is  the  ratio  of  a  to  e  ? 
Ex.  3.   If  a  +  e  :  a  —  e  : :  7  :  5,  what  is  the  ratio  of  a  to  e  ? 

What  is  the  ratio  of  x  and  y  in  the  following : 
Ex.  4.  x  +  y:x::lS:i? 

Ex.  5.  x-y:x::6:9? 

Ex.  6.  x-hy:x-y::lZ:6? 

Give  the  name  and  value  of  x  in  each  of  the  following  proportions : 
Ex.  7.  5  :  a; : :  10  :  12. 

Ex.  8.  a; :  8  : :  10  :  16. 

Ex.  9.  5  :  11  : :  10  :  a;. 

What  is  the  name  and  what  the  value  of  x  in  the  following : 
Ex.  10.  4  :  X  : :  a  :  36  ? 

Ex.11.  rB:8::8:2? 

Give  ten  proportions  that  can  be  derived  from  the  proportion : 

Ex.  12.  3  :  7  : :  9  :  a:. 

Ex.  13.  Test  the  correctness  of  your  answer  by  showing  that  the 
product  of  the  means  and  extremes  in  the  derived  proportions  is  identical 
with  the  product  of  the  means  and  extremes  in  the  original  proportion. 


108  THE   ELEMENTS  OF  GEOMETRY 

Method  op  Limits 

DEFINITIONS 

A  Variable  is  a  quantity  that  in  the  course  of  a  single  dis- 
cussion is  always  changing  its  value. 

Thus,  as  the  point  X  moves  along  the  curve 
ABy  its  distance  from  the  line  AB,  its  distance 
from  the  point  A^  and  the  projection  AY,  of 
this  latter  distance  on  the  line  AB,  are  all 
variables. 

A  Constant  is  a  quantity  that  does  not  change  its  value  in  the 
course  of  a  single  discussion. 

Thus,  if  the  curve  in  the  above  figure  be  a  semicircle  on  AB  as  a  di- 
ameter, AB  is  a  constant.  The  changes  in  the  variables  above  mentioned 
produce  no  changes  in  AB. 

A  Limit  of  a  variable  is  a  constant  which  the  value  of  the 
variable  may  be  made  to  approach  as  near  as  we  please,  but 
which  the  variable  cannot  be  made  to  reach.  That  is,  the  dif- 
ference between  the  limit  and  the  variable  may  be  made  less 
than  any  assignable  quantity,  but  cannot  be  made  zero. 


A  G  E    F    B 

To  illustrate  :  Suppose  a  point  moves  from  A  toward  jB  so  as  to  cover 
one  half  the  distance  in  the  first  second,  one  half  the  remaining  distance 
in  the  second,  and  so  on. 

Will  the  moving  point  ever  coincide  with  B  ? 

In  other  words,  will  the  variable  distance  covered  by  the  moving  point 
ever  coincide  with  the  constant  line-segment  AB? 

What,  then,  is  the  Limit  of  the  variable  distance  gone  over  by  the 
moving  point  ? 

If  the  distance  passed  over  in  the  first  second  be  called  1,  that  passed 
over  in  the  second  second  will  be  \,  that  passed  over  in  the  third  second 
will  be  i,  and  so  on. 

The  whole  distance,  therefore,  say  x,  will  be  l  +  i  +  i  +  J+T^+^jH — • 

The  greater  the  number  of  terms  we  take,  the  nearer  x  will  approach 
the  value  2. 


XL     MEASUREMENT  109 


1  +  J  +  i              =  1; 

2-   1=  i 

i+i+i+i        =¥; 

2-¥=  i 

i  +  i  +  i  +  i  +  i^  =  fi; 

2  -  fi  =  tV 

Thus, 


and  so  on. 

We  can  make  x  as  near  2  as  we  please  ;  i.e.  differ  from  2  by  as  small  a 
fraction  as  we  choose  by  taking  a  sufficiently  large  number  of  terms. 

But,  no  matter  how  great  the  number  of  terms  we  take,  their  sum  will 
never  actually  reach  2. 

/.  2  is  said  to  be  the  limit  of  the  sum  of  the  terms. 

And  AB  is  the  limit  of  the  sum  of  the  segments  J.C,  CJS',  EF^  etc.  ;  i.e, 
the  Limit  of  the  Variable  Distance  gone  over  by  the  moving  point  is  AB. 

The  symbol  =  is  employed  to  denote  the  expression  "ap- 
proaches as  a  limit,"  or  any  equivalent  expression. 

Postulate  of  Limits.  If,  while  approaching  their 
respective  limits,  two  variables  are  alivays  equal,  the 
limits  are  equal. 

For,  since  the  two  variables  are  equal  at  every  stage  of  their  progress, 
we  have  practically  but  one  variable  ;  and  it  is  impossible  that  one  vari- 
able (increasing  or  decreasing)  should  be  approaching  two  different  limits 
at  the  same  time. 

Direct  inferences : 

(a)  The  limit  of  the  pi*oduct  of  two  variables  is  the  product 

of  their  limits. 

(b)  If  two  variables  have  a  constant  ratio,  and  each  ap- 

proaches  a   limit,  these  limits,  taken   in   the    same 
order,  have  the  constant  limit  of  the  ratios. 

Ex.  14.  25  and  49  are  perfect  squares.  By  what  proposition  does  it 
follow  that  their  product  must  be  a  perfect  square  ? 

Ex.  15.  27  and  125  are  perfect  cubes.  By  what  proposition  does  it 
follow  that  3375  is  also  a  perfect  cube? 

Ex.  16.  Verify  all  the  conclusions  of  XL  1,  Sch.  2  by  use  of  the  test 
given  in  XL  1,  Sch.  1. 

Ex.  17.    Show  that,  if     a  :  6  :  :  6  :  c  : :  c  :  e, 

then  a  :  e :  :  a*  :  6*. 


110  THE  ELEMENTS  OF  GEOMETRY 


XI.     SUMMART  OF  PROPOSITIONS   IN   THE  GROUP 
ON  MEASUREMENT 

1.  If  four  quantities  are  in  'proportion^  the  product 
of  the  means  equals  the  product  of  the  extremes,  and 
conversely, 

2.  In  any  number  of  equal  ratios,  the  sum  of  the 
antecedents  is  to  the  sum  of  the  consequents  as  any 
antecedent  is  to  its  consequent. 

3.  If  two  proportions  he  midtiplied  together,  term  hy 
term,  the  resulting  products  will  be  in  proportion. 

a   If  any  number  of  proportions  he  midtiplied 

together,  term  by  term,  the  resulting  products 

will  be  in  proportion, 
b   Like  powers  of  the  terms  of  a  proportion  are 

in  proportion. 
e   Like  roots  of  the  terms  of  a  proportion  are  in 

proportion, 

4.  If  the  terms  of  a  proportion  he  divided  successively 
hy  the  terms  of  a  secojid,  the  resulting  quotients  will  he 
in  proportion. 

Postulate  of  Limits.  If,  while  approaching  their 
respective  limits,  two  variables  are  always  equal,  the 
limits  are  equal. 


XII.  GROUP  ON  MEASUREMENT  OF  ANGLES 

DEFINITIONS 

A  Central  Angle  is  an  angle  whose  vertex  is  at  the  center  of  a 
circle. 

An  Inscribed  Angle  is  an  angle  formed  by  two  chords  inter- 
secting on  the  circumference. 

An  Escribed  Angle  is  an  angle  formed  by  one  chord  with 
another  chord  produced. 

An  angle  is  inscribed  in  a  segment  when  its  vertex  is  in  the 
arc  of  the  segment  and  its  sides  pass  through  the  extremities 
of  this  arc. 

The  angle  between  two  curves  at  any  point  of  intersection  is 
the  angle  formed  by  the  tangents  to  the  curves  at  this  point. 
If  the  angle  between  two  curves  is  right,  the  curves  are  said 
to  cut  each  other  orthogonally. 


Thus,  a  AT,  BT  (Fig.  1,  Fig.  2)  be  tangent  to  the  circles  K  and  i, 
respectively,  Z.ATB  is  the  angle  between  the  circles  at  T.  If,  as  in  Fig.  2, 
Z  A  TB  is  right,  the  circles  are  said  to  cut  each  other  orthogonally. 

The  angle  KTL,  between  the  radii  to  T,  is  supplemental  to  Z.ATB 
(IX,  1  and  II,  5).  Hence,  \i  /.ATB  is  right,  Z  KTL  is  right,  and  con- 
versely, that  is, 

Two  circles  cut  each  other  orthogonally  when  the  radii  to 
either  point  of  intersection  are  perpendicular  to  each  other. 

Ill 


112 


THE  ELEMENTS  OF  GEOMETRY 


PROPOSITIONS 

XII.  1.  In  the  same  circUy  or  equal  circles,  central 
angles  are  to  each  other  as  their  intercepted  arcs. 

Hyp.  Case  I.  If 
O  A^  =  O  /li,  and 
arc  AB  is  to  arc  CE 
as  any  two  whole 
numbers,  say  6  to 
4   (commensurable), 

Cone. :  then  Z  AKB :  /.  CK^E : :  arc  AB :  arc  CE. 

Dem.     Arcs  AB  and  CE  have  a  common  measure.       (Hyp.) 
Apply  it  to  each  of  the  arcs,  and  suppose  it  is  contained 

six  times  in  AB  and  four  times  in  CE. 

* 

Join  the  points  of  division  with  the  centers  of  the  circles. 
All  the  central  angles  thus  formed  will  be  equal.  (IX.  3.) 

.-.  Z^7fJ3:ZC/A^::6:4. 

But  arc^l^rarcO^    : :  6  :  4. 

.-.  Z  AKB.  A  CK^E'.'.  arc  ^5:  arc  CE. 

Q.E.D. 


Hyp.   Case  II.  If 

Oir=0^i,  and 
arc  AB  and  arc  CE 
are  incommensura- 
ble, 


Cone. :  then  Z  AKB :  Z  CK^E  : :  arc  AB :  arc  CE. 

Dem.  Divide  arc  AB  into  any  number  of  equal  parts,  say  4, 
each  less  than  arc  CE.  Let  AM  be  one  of  these  parts,  and  be 
contained  in  CE  twice,  with  a  remainder  LE. 

As  the  remainder  is  always  less  than  the  divisor,  it  follows 
that  if  we  increase  the  number  of  equal  parts  into  which  AB 
is  divided,  we  diminish  both  the  divisor  and  the  remainder. 


XII.     MEASUREMENT  OF  ANGLES  113 

But,  as  the  arcs  AB  and  CE  are  incommensurable  (Hyp.), 
the  remainder  can  never  be  0. 

,\  arc  CL  =  arc  CE,  and  Z  CK^L  =  Z  CK,E. 

(XI.  Def.  of  Limit.) 

^"*  44^  always  =  ^^^>       (XII.  1,  Case  I.) 

Z.  (JKiL  arc  iJL 

.-.  Z  ^^5 :  Z  CK^E  : :  arc  ^5 :  arc  CE.     (XI.  Post.  Limits.) 

Q.E.D. 

ScH.  Heretofore;  we  have  measured  angles  directly  (see  p. 
100),  using  the  right  angle  as  the  unit  of  measure.  This  unit 
is  inconvenient,  however,  as  its  use  requires  us  to  employ 
fractions  too  frequently. 

The  foregoing  theorem,  due  to  Thales  of  Miletus  (640  b.c), 
introduced  a  very  simple  method  of  indirect  measurement. 

Any  change  in  the  magnitude  of  the  central  angle  produces 
a  proportional  change  in  the  intercepted  arc.  Thus,  if  the  cen- 
tral angle  be  doubled,  the  intercepted  arc  is  doubled;  if  the 
angle  be  trebled,  the  arc  is  trebled ;  and  so  on. 

Hence,  the  intercepted  arc  may  be  taken  as  the  measure  of 
the  central  angle. 

This  very  important  theorem  may  be  stated  thus : 

A  central  angle  is  measured  by  its  intercepted  arc. 

The  circumference  is  divided  into  360  equal  parts,  called 
degrees ;  each  degree  into  60  equal  parts,  called  minutes,  and 
each  minute  into  60  equal  parts,  called  seconds.  For  brevity, 
an  angle  measured  by  an  arc  of  1°  is  called  an  angle  of  1° ;  etc. 

Note.  — This  division  of  the  circumference,  known  as  the  sexagesimal 
division  (or  division  by  sixties) ,  is  due  to  the  Babylonians. 

The  Babylonian  year  consisted  of  12  months  of  30  days  each,  or  360 
days.  Accordingly  the  zodiac  was  divided  into  12  signs  of  30  degrees 
each,  making  360  degrees  for  a  complete  circle. 

These  people  were  also  familiar  with  the  fact  that  the  radius  of  a 
circle,  used  as  a  chord,  divides  the  circle  into  6  equal  parts  of  60  degrees  * 
each.     Hence,  arose  the  custom  of  subdividing  the  degree  by  sixties,  into 
minutes,  etc.,  as  indicated  above. 


114 


THE  ELEMENTS  OF  GEOMETRY 


XII.  2.  An  inscribed  angle  is  measured  by  half  the 
intercepted  arc. 


Case  (1) 
B 


Case  (2) 

li 


Hyp.  If  Z  5  is  inscribed  in  OKy  and  if  AC  is  the  inter- 
cepted arc, 

Cone:  then  Z^  is  measured  by  E^jd^. 

^        2 

Case  (1).  When  one  arm  of  the  angle  is  a  diameter  through  B. 

Case  (2).  When  the  two  arms  are  on  opposite  sides  of  the 
diameter  through  B. 

Case  (3).  When  both  arms  are  on  the  same  side  of  the  diam- 
eter through  B. 

Dem.    (1)  Draw  the  radius  CK,  forming  the  isosceles  A  CKB. 

•Then  ZB  =  Z.a  (IV.  L) 

.-.  ZB  =  :^/.AKa  (III.  2  a.) 

But  Z  AKC  is  measured  by  arc  AG,  (XII.  1.  Sch.) 

.*.  Z  jB  is  measured  by  \  arc  AC. 

Dem.     (2)  Draw  the  diameter  BM. 

Then  Z  MBC  is  measured  by  |  arc  MC.      (XII.  2,  Case  (1).) 

Z  ABM  IB  measured  by  \  ^.tgAM.  (XII.  2,  Case  (1).) 

.-.  Z  MBC  +  Z  ABM  is  measured  by  ^  arc  MC  +  ^  arc  AM. 

.'.  Z  ABC  is  measured  by  4  arc^C. 

Q.E.D. 

Dem.     (3)  Both   arms   on  the   same  side  of  the  diameter 
through  B. 
Proof  to  be  supplied  by  the  pupil. 


Q.E.D. 


XII.     MEASUREMENT  OF  ANGLES  115 

XII.  2  a.  An  angle  formed  by  a  tangent  and  a  chord 
is  measured  by  half  the  intercepted  arc. 


Hyp.    If  AO 

and  EF  inter- 
sect at  C,  and 
if  ^Ois  a  chord 
and  EF  is  a  tan- 
gent, 


Cone. :  then  Z  AOF  is  measured  by 


2 

Dem.     Draw  the  diameter  CL. 
Z  LCF  is  a  right  angle. 

[A  radius  to  point  of  tangency  A.  the  tangent.]  (IX.  4.) 

ZACF=Tt.Z.LCF+ZLCA.  (Ax.  4.) 

Kt.  Z  LCF  is  measured  by  ESJ^, 

[A  right  angle  is  measured  by  one  half  a  semicircumference.] 

(XII.  1,  Sch.) 

/LLC A  is  measured  by   ?^^^.  (XII.  2.) 

.-.  Z^OT  is  measured  by  arc  XmC     arc  ^i^  ^  arc  ^mC 

•^2^2  2 

Q.E.D. 

ScH.     Angles  inscribed  in  the  same  segment  are  equal. 

If  the  segment  is  greater  than  a  semicircle,  the  angles  are 
acute. 

If  the  segment  is  a  semicircle,  the  angles  are  right. 

If  the  segment  is  less  than  a  semicircle,  the  angles  are 
obtuse. 


Ex.  1.  The  bisectors  of  the  vertex  angles  of  all  triangles  on  the  same 
base  and  inscribed  in  the  same  segment  are  concurrent. 


116         THE  ELEMENTS  OF  GEOMETRY 

XII.  3.  An  angle  ivhose  vertex  lies  between  the  center 
and  the  circumference  is  measured  by  half  the  sum  of 
its  intercepted  arcs. 

Hyp.     If  AC  and 

BE  intersect  at  V 
lying  between  the 
center  and  the  cir- 
cumference, 

Cone. :  then  ZAVB  is  measured  by  arc ^g  +  arc  C^^ 

Dem.     Draw  BC. 

ZAVB  =  ZC+ZB. 

[The  ext.  Z  of  a  A  equals  the  sum,  etc.]  (III.  2.) 

But  Z  C  is  measured  by  ^:E£AM.  (XII.  2.) 

And  Z  7?  is  measured  by  ^^^/^^.  (XII.  2.) 


2 
arc  AB  +  arc  CE 


Z  A  VB  is  measured  by 


Q.E.D. 


Ex.  2.    The  opposite  angles  of  an  inscribed  4-side  are  supplemental. 

Ex.  3.  The  bisector  of  any  interior  angle  of  an  inscribed  4-side  and 
the  bisector  of  the  opposite  exterior  angle  intersect  on  the  circum- 
ference. 

Ex.  4.  How  many  degrees  are  there  in  the  arc  that  subtends  an  in- 
scribed angle  of  25°  ?     Of  25°  40'?    Of  25°  40' 84"? 

Ex.  5.  What  arc  measures  the  supplemental  adjacent  angles  of  the 
preceding  inscribed  angles  ? 

Ex.  6.    What  are  these  angles  called  ? 

Ex.  7.  An  angle  between  the  center  and  the  circumference  is  40°  30'. 
What  is  the  sum  of  the  arcs  that  measure  it  ? 

Ex.  8.  An  angle  of  70°  and  its  supplement  are  formed  by  a  tangent 
and  a  chord.     What  is  the  value  of  each  arc  that  subtends  these  angles  ? 

Ex.  9.  The  arc  of  a  segment  is  140°.  What  is  the  value  of  each  angle 
inscribed  in  this  segment  ? 


_<^ 


XII.     MEASUREMENT   OF  ANGLES  117 

XII.  4.  An  angle  formed  hy  two  secants  intersecting 
without  the  circle  is  measured  hy  half  the  difference  of 
the  intercepted  arcs. 

Hyp.    If  AQ 

and  BE  inter- 
sect at  V  lying 
without  the  cir- 
cumference, and 
AC  and  BE  are 
both  secants, 

Cone. :  then  ZAVBi^  measured  by  arc  ^^ -- arc  (7^^ 

Dem.     Draw  BC. 

AACB  =  ZAVB  +  ZB.  (1.) 

[The  ext.  Z  of  a  A  equals  the  sum,  etc.]  (III.  2.) 

.-.  AAVB  =  ZACB~ZB. 

((1.)  by  transposition.) 
But  Z  ACS  is  measured  by  H2^.  (XII.  2.) 

ZBis  measured  by  ^^5-^.  (XII.  2.) 

.'.  ZAVB  is  measured  by    . 

arc  AB     arc  CE  _  arc  AB  —  arc  CE 

2  2      ~  2        ■      * 

Q.E.D. 

Prove  by  means  of  XII.  2  : 

Ex.  10.    That  an  isosceles  triangle  is  isoangular. 

Ex.  11.    That  an  isoangular  triangle  is  isosceles. 

Ex.  12.    That  the  sum  of  the  interior  angles  of  a  tri- 
*igle  equals  two  right  triangles. 

Ex.  13.    That  the  sum  of  the  exterior  angles  of  a 
triangle  equals  four  right  angles. 

Ex.  14.  That  two  mutually  equiangular  triangles  inscribed  in  the  same 
circle  are  congruent. 


118 


THE  ELEMENTS  OF  GEOMETRY 


XII.  4  a.  An  angle  formed  hy  a  tangent  and  a  secant 
is  measured  hy  half  the  difference  of  the  intercepted 
arcs. 


Hyp.  liAQ 
is  a  secant  and 
BE  is  a  tangent 
(XII.  4  a),  or  if 
both  are  tan- 
gents (XII.  46), 


Fig.  2. 


Cone. :  then  Z  AVB  is  measured  by 


Fig.  3. 
arc  AB  —  arc  CE 


Dem.     Similar  to  Demonstration  of  XII.  4. 

(Let  the  pupil  give  it  in  full.  Note  that  when  the  secant  VB  of  Fig.  1 
is  turned  on  F  as  a  pivot  until  it  becomes  a  tangent,  the  points  B  and  E 
become  coincident,  as  shown  in  Fig.  2,  and  may  be  denoted  by  a  single 
letter.  Observe  also  that  if  the  secant  VA  of  Fig.  1  be  likewise  turned  on 
F  as  a  pivot  until  it  becomes  a  tangent,  the  points  A  and  C  of  Fig.  1 
become  coincident,  as  shown  in  Fig.  3,  and  may  be  denoted  by  a  single 
letter.'i 

XIT.  4  h.  An  angle  formed  hy  tivo  tangents  is  meas- 
ured hy  half  the  difference  of  the  intercepted  arcs. 


Ex.  15.  The  4-side  ABCE  is  inscribed  in  a 
circle.  Zi^  is  28°;  ZBOC  is  82^  How  many 
degrees  in  each  of  the  arcs  CB  and  AE  ? 

How  many  degrees : 

Ex.  16.    In  Z  OBF  &ndZOCF? 

Ex.  17.    In  ZBAE  and  ZBAC? 

Ex.  18.    In  Z  GEO  ? 

Ex.  19.  What  kind  of  angle  with  reference  to 
the  circle  is  Z  GEO  ? 

Ex.  20.  What  is  the  value  of  all  the  angles 
that  are  inscribed  in  the  major  segment  that 
stands  on  the  chord  AE  ? 


XII.    MEASUREMENT  OF   ANGLES  119 


XII.     SUMMARY  OF  PROPOSITIONS  IN   THE  GROUP 
ON  MEASUREMENT   OF  ANGLES 

1.  In  the  same,  or  equal  circles,  central  angles  vary 
{are  to  each  other)  as  their  intercepted  arcs. 

ScH.    A  central  angle  is  measured  by  its  intercepted  arc. 

2.  An  inscribed  angle  is  measured  by  half  the  inter- 
cepted arc. 

a  An  angle  formed  by  a  tangent  and  a  chord  is 
measured  by  half  the  intercepted  arc. 

3.  An  angle  ivhose  vertex  lies  between  the  ceiiter  and 
the  circumference  is  measured  by  half  the  sum  of  its 
intercepted  arcs. 

4.  All  angle  formed  by  two  secants  intersecting  ivith- 
out  the  circle  is  measured  by  half  the  difference  of  the 
intercepted  arcs. 

a  An  angle  formed  by  a  tangent  and  a  secant 
is  measured  by  half  the  difference  of  the 
intercepted  arcs. 

b  An  angle  formed  by  two  tangents  is  measured 
by  half  the  difference  of  the  intercepted  arcs. 


120         THE  ELEMENTS  OF  GEOMETRY 

Hints  to  the  Solution  of  Original  Exercises 
In  solving  a  problem  in  algebra  we  proceed  as  follows : 

1.  We  assume  that  we  have  the  quantity  required  and  call  it  x. 

2.  We  form  an  equation  in  which  x  may  be  surrounded  by  a  number 
of  modifiers — coefficients,  exponents,  etc. 

The  value  of  x  is  found  when  by  a  transformation  or  by  a  series  of 
transformations,  x  stands  alone  on  one  side  of  the  equation^  while  the 
modifiers  in  some  form  appear  on  the  other. 

In  solving  a  problem  in  geometry  we  proceed  in  a  similar  way  : 

1.  We  assume  that  we  have  the  figure  that  satisfies  the  conditions 
given  in  the  problem.  This  assumed  figure  corresponds  to  the  x  of 
algebra. 

2.  We  ask  ourselves  what  follows  from  this  assumption  ;  that  is,  what 
definitions,  axioms,  or  previously  established  theorems,  corollaries,  or 
problems  are  suggested  by  the  assumed  figure. 

.3.  We  ask  what  one  of  these  theorems  or  combination  of  them  may  be 
applied  in  the  actual  construction  of  the  required  figure. 

These  applied  propositions  correspond  to  the  modifiers  ofx  in  algebra. 

The  drawing  of  such  auxiliary  line  or  lines  as  will  make  it  possible  to 
apply  a  suggested  theorem,  or  a  combination  of  suggested  theorems,  as  well 
as  the  discovery  of  these  theorems,  is  the  test  of  the  inventional  power  of 
the  student.  No  rule  can  be  made  that  will  tell  him  what  theorem  to 
select  or  what  line  to  draw,  but  the  systematic  attack,  persistently  made, 
familiarizes  him  with  the  principles  of  geometry. 

The  solution  of  a  problem  consists  of : 

1.  The  analysis  as  outlined  above.  • 

2.  The  construction. 

3.  The  proof. 

4.  The  discussion. 

Many  problems  in  the  beginning  of  the  course  in  algebra  may  be  solved 
without  the  use  of  x.  That  is,  they  may  be  considered  problems  in 
arithmetic. 

So,  in  geometry,  it  is  by  no  means  always  necessary  to  give  the  analysis 
of  the  problem  ;  that  is,  to  assume  we  have  the  required  figure,  etc.  We 
pass,  however,  from  the  simple  to  the  complex.  We  learn  best  how  to 
use  the  method  in  problems  where  it  is  indispensable  by  applying  it  to  the 
solution  of  simpler  problems  first. 


XII.     MEASUREMENT  OF  ANGLES 


121 


PROBLEMS,   EXERCISES,    AND   SPECIAL   THEOREMS 
Problems 

Prob.  I.    O71   a   given   line   to   construct   a   circular 
segment  which  shall  contain  a  given  angle. 


Given.     A 

line-segment 
AB  and  an 
angle  E. 


Required.  On  AB  as  a  chord  to  construct  a  circular  segment 
capable  of  containing  an  angle  equal  to  Z  E. 

Const.     Construct  an  Z  FRH=  Z  E.  (V.  Prob.  III.) 

With  AB  as  a  radius,  and  any  point  J  in  BF  as  a  cen- 
ter so  taken  that  the  arc  described  will  cut  RH,  describe 
an  arc. 

Let  it  intersect  BH  in  Q, 

Draw  QJ,  and  draw  a  circle  through  B,  Q,  and  J. 

(Prob.  IX.,  Ex.  in  drawing,  p.  17.) 

The  segment  QRJ  is  the  segment  required. 

Q.E.D. 

Dem.  Any  angle  inscribed  in  this  segment  is  measured  by 
one  half  the  same  arc  that  subtends  Z  FRH.  (XII.  2.) 


all  such  angles  equal  Z.FBH—Z.E. 


Q.E.D. 


A  second  construction  in  common  use  is  the  following : 
Erect  a  mid  ±  to  AB.     At  B  make  Z  ABC  =  Z  E,  and  erect 
BMA.BC.     The  O  with  K,  the  intersection  of  these  Js,  as 
center  and  KB  as  radius  gives  the  required  segment. 


122 


THE  ELEMENTS  OF  GEOMETRY 


Illustration  of  the  Method  of  Solving  Original 
Problems 

Prob.  II.     Gfiven   the   base,  vertex  angle,  and  sum 
of  the  legs  of  a  triangle,  construct  it. 


Given.     The  afb 

c,  the 
vertex  Z  C, 
and  the  sum 
a  +  b. 


Required.  A  triangle  whose  base  equals  c,  vertex  angle  equals 
Z  C,  and  the  sum  of  the  legs  equals  a  -\-b. 

Analysis.  Assume  A  ABC  has  c'  =  c,  Z.C  =  Z.C,  and 
a' H- 6' =  a -f- 6. 

It  follows  that  if  AC  be  produced,  making  CB'  =  a',  and 
if  BB'  be  drawn,  A  CB'B  is  isosceles. 

Suggested  theorems  are :  III.  2  a  and  IV.  1 ;  also  Axioms 
7  and  5. 

Applicable  theorems,  etc. :   All  the  above;  for  the  ZJB'  and 

(IV.  1  and  III.  Ex.  4.) 

Z.  C 
.'.  if  we  start  with  a +  6,  since  ZJ5'  =  -— -,  the  position  of 

B'B  is  known.  ^  (Ax.  7.) 

And,  since  5  is  c  distant  from  A  and  also  somewhere  in  BB' 

or  its  extension,  B  is  known.  (Ax.  5.) 

Similarly  the  position  of  BC  is  known.  (Ax.  7.) 

Hence,  C  is  known.  (Ax.  5.) 

.*.  we  have  discovered  how  to  construct  the  required  triangle. 


the  Z  CBB'  each  equal  — • 


Note.  —  An  analysis  when  complete  shows  us  clearly  the  method  of 
construction.  In  short,  the  construction  then  becomes  merely  an  exercise 
in  geometrical  drawing. 


XII.     MEASUREMENT  OF  ANGLES  123 

Const.     Let  AB'  =  a  +  6.  :^ 

At  B'  construct  an  angle  with  AB'  ==  — —•  /  \ 

With  ^  as  a  center  and  a  radius  equal  to  c,  /^n\^ 

describe  arc  BB".  G't     J/^^ 

Draw  BO  making  Z  B'BG^  Z.  B\  /^^^\ 

Draw  AB.  ^r^.___>4 

A  ^5(7  is  the  required  triangle. 

Q.E.F. 

Proof.  ZB'BG=ZB'  =  ^'  (Construction.) 

/.  A  B'BC  is  isoangular. 

.-.  ZACB  =  2ZB'. 

[The  ext.  vert.  Z  of  an  isoangular  A  equals,  etc.]    (III.  2  a.) 

.-.  Z^(75  =  givenZa  (1) 

Again,  B'G=Ba 

[An  isoangular  triangle  is  isosceles.]  (IV.  2.) 

.-.  AC+BC=AG+B'G=a  +  b.  (2) 

AB  =  c.     (Construction.)  (3) 

Q.E.D. 

Discussion.  A  second  triangle  fulfilling  the  given  conditions 
may  be  formed ;  for  the  circle  of  line  (3)  of  the  above  construc- 
tion cuts  B'B  in  a  second  point,  B".  Therefore,  draw  B"G' 
just  as  -BC  was  drawn  and  draw  the  join  B"A. 

A  AB"0'  is  the  second  triangle  meeting  the  given  conditions. 

If  the  circle  with  ^  as  a  center  and  c  as  a  radius  is  tangent 
to  BB',  the  two  triangles  coincide  and  become  identical. 

If  this  circle  does  not  cut  BB',  no  such  triangle  can  be  con- 
structed. 

Exercises 

The  following  exercises  are  on  the  loci  of  vertices  of  the 
equal  vertex  angles  of  triangles  standing  on  the  same  base; 
of  the  incenters  of  such  triangles ;  of  their  excenters,  circum- 
centers,  orthocenters,  and  centroids. 


124 


THE  ELEMENTS  OF  GEOMETRY 


Ex.  a.  Hyp.  If  any  number  of 
angles  equal  Z  C,  and  their  sides  pass 
through  the  ends  of  the  given  line- 
segment  ABy 


Cone. :  then  the  locus  of  their  vertices  is  a  circular  segment  on  AB  as 
a  chord,  and  capable  of  containing  an  angle  equal  to  Z  C. 

Demi.    On  AB,  equal  to  the  given  line,  as  a  chord,  construct  a  circular 
segment  capable  of  containing  an  angle  equal  to  the  given  Z  C. 

(Prob.  I.  Group  XII.) 
The  arc  of  this  segment  is  the  locus  required.     For, 


(1)  Any  angle  inscribed  in  this  segment  equals  Z  C, 
-:  it  is  measured  by  H^A^. 


(XII.  3.) 


(2)  No  angle  whose  vertex  is  without  the  circle,  subtended  by  AB  and 
to  the  left  of  it,  can  equal  Z  C.     For, 

Draw  the  auxiliary  HB.   Z  AHB  {=  ZC.    Why  ?)  is  greater  than  Z  V. 

[The  ext.  Z  of  a  A  >  either  non-adj.  int.  Z.]  (III.  2,  Sch.) 

Similarly  it  may  be  shown  that  an  angle  whose  vertex  lies  within  the 
circle  cannot  equal  Z  C. 

.'.  the  arc  of  the  segment  to  the  left  of  AB,  on  AB  as  a  chord,  is  the 
locus  of  the  vertices  of  all  angles  to  the  left  oi  AB  subtended  by  AB. 

The  locus  of  the  vertices  of  such  angles  on  the  right  of  AB  and  equal 
to  ZC  will  be  the  arc  of  a  segment  on  AB  as  a  chord  on  the  right  of  AB, 
and  equal  to  that  on  the  left. 

Q.E.D. 

Note  1. — If  the  given  angle  be  acute,  both 
arcs  will  be  major. 

If  the  given  angle  be  right,  both  arcs  will  be 
semicircles. 

If  the  given  angle  be  obtuse,  both  arcs  will  be 
minor. 

Note  2.  —  The  pupil  will  remember  that  in  order  to  prove  that  a  line  is 
the  locus  of  a  point,  two  things  must  be  established. 
What  are  they  ? 

Ex.  b.   The  locus  of  the  incentere  of  triangles  whose 
base  is  AB  and  whose  vertex  angles  always  =  Z  C,  is  : 
The  arc  of  a  segment  on  AB  as  a  chord  that  will 
ZC^ 
2  * 


contain  an  Z  =  rt.  Z  + 


XII.     MEASUREMENT   OF   ANGLES 


125 


Proof.     Z  AKiB  =  1  rt.  Z  +  ^. 


(IV.  Ex.  20.) 


.'.  by  XII.  Ex.  (a)  the  locus  of  Kt  is  the  arc  of  a  segment  on  AB  as  a 

Q.E.D. 


chord  that  will  contain  an  angle  =  rt.  Z  ^ 


Ex.  c.   The  locus  of  the  excenters  of  triangles  whose  base  is  AB  and 
whose  vertex  angles  always  equal  Z  (7,  is  : 

The  arc  of  a  segment  on  AB  as  a  chord  that  will  contain  an  Z = rt.  Z 


Proof.  KeA  and  KgB  bisect  ext.  Z  A  and 
ext.  Z  B,  respectively. 

Draw  KiA  and  KiB  bisecting  int.  Z  A  and 
int.  Z  J5,  respectively. 

Z  iTi  =  rt.  Z  +  — .      (v.  preceding  Ex.  6.) 

Z  KiAKe  and  Z  KiBK^  are  right  angles.    Why  ? 
.-.  Z  iTi  +  Z  /Q  =  2  rt.  A. 

2 


.'.  the  locus  of  Ke  is  the  arc  of  a  segment  on  AB  as  a  chord  that  will 
ZC_ 

2 


contain  an  Z  =  rt.  Z 


Q.E.D. 


Ex.  d.   The  locus  of  the  circumcenters  of  triangles  whose  base  is  AB 
and  whose  vertex  angles  always  equal  Z  (7,  is  : 

The  arc  of  a  segment  on  AB  as  a  chord  that  will  contain  an  Z  =  2  Z  C. 


Proof.  Kc  is  by  hypothesis  the  center  of  the  cir- 
cumcircle  of  triangle  ABC.  (v.  X.  3.) 

.'.  Z  AKcB  is  a  central  angle  and  is  measured  by 
the  arc  AB.  (XII.  1,  Sch.) 

But  Z  C  is  inscribed  in  this  circle. 

.".  Z  C  is  measured  by  half  the  arc  AB. 

:.  ZAKcB  =  2ZC. 

:.  the  locus  of  Kc  is  the  arc  of  a  segment  on  AB  as  a  chord  that  will 

contain  an  angle  equal  2ZC. 

Q.E.D. 


126 


THE  ELEMENTS  OF  GEOMETRY 


Ex.  «.  The  locus  of  the  orthocentera  of  triangles  whose  base  is  AB 
and  whose  vertex  angles  always  equal  Z  C,  is  : 

The  arc  of  a  segment  on  AB  as  a  chord  that  will  contain  an  angle  equal 
to  the  supplement  of  Z.  C. 

Proof.  •  Z  AKoB  =  /.  LKo  T.  ^ 

•/  Z  CLKo  is  a  right  angle  and  Z  CTKo  is  a  right 
angle,  Z  C  is  the  supplement  of  Z  LKoT. 

:.  Z  C  is  the  supplement  of  Z  AKoB. 

:.  the  locus  of  Ko  is  the  arc  of  a  segment  on  AB  as 
a  chord  that  will  contain  an  angle  equal  to  the  supplement  of  Z  C. 


Ex.  /.  If  the  opposite  angles  of  4-side  ABCE  are 
supplemental,  show  that  a  circle  passing  through  A,  B, 
and  C,  will  also  pass  through  E ;  that  is,  if  the  oppo- 
site angles  of  a  4-side  are  supplemental,  it  is  a  cyclic. 


Ex.  g.   Let  the  student  state  the  locus  of  each  center  given. 


Ex.  21.  Describe  a  circle.  Draw  a  4-side  so  that  one  of  its  angles  shall 
be  formed  by  two  tangents,  one  Jsy  a  tangent  and  a  chord,  one  by  a  tangent 
and  a  secant,  and  the  fourth  by  two  secants. 

Point  out  the  measure  of  each  of  the  above  four  angles. 


Ex.  22.  If  two  chords  are  perpendicular  to  each 
other  in  a  circle,  the  sum  of  either  set  of  opposite 
intercepted  arcs  is  a  semicircle. 


Ex.  23.  If  at  the  vertex  of  an  inscribed 
equilateral  triangle  a  tangent  is  drawn,  find 
the  angle  between  the  tangent  and  each  of 
the  sides  meeting  at  the  vertex. 


Ex.  24.  Tangents  are  drawn  at  the  vertices  of  an  inscribed  triangle. 
Two  angles  of  the  inscribed  triangle  are  respectively  70°  and  80°.  Find 
the  angles  of  the  circumscribed  triangle. 


XII.     MEASUREMENT   OF   ANGLES 


127 


Theorems  of  Special  Interest 
Pedal  Triangle 


Th.  1.  If,  in  A  ABC, 
P,  E,  and  D  are  the  feet 
of  the  altitudes  on  c,  a, 
and  h,  respectively, 


A  F  <i  B 

Cone:  then  CP bisects  ADPE,  AE  bisects  Z Z>J5;P,  and  DB 
bisects  Z  EDP. 


Dem.     CP,  BD,  and  AE  concur  at  K„. 


(X.  4.) 


The  4-side  PKoDA  is  cyclic. 

[If  the  opposite  Z  of  a  4-side  be  supp.,  it  is  cyclic] 

(Ex./,  p.  126.) 
.-.  Z.DPIC  =  Z.DAK., 


(XII.  3.) 


Both  are  measured  by  ^^^-- — -- 

The  4-side  APEC  is  cyclic. 

[Rt.  A  CEA  and  CPA  stand  on  the  same  hypotenuse  AC.^ 

.-.  Z.DAK.=  Z.CPE. 


Both  are  measured  by 


arc  0^  n 

^       J 


A  CPE. 


.'.  ZDPK, 

.\  CP  bisects  Z  DPE. 

Similarly,  AE  bisects  ZD^P  and  DB  bisects  Z.PDE 

Note.  — APJED  is  known  as  the  pedal  triangle  of  A  ABC. 


(XII.  3.) 
(Ax.  1.) 


Q.E.D. 


Ex.  25.  If,  through  the  point  of  con- 
tact of  two  circles  tangent  externally,  a 
straight  line  is  drawn  terminating  in  the 
circles,  the  tangents  at  the  extremities  of 
this  line  are  parallel. 


128         THE  ELEMENTS  OF  GEOMETRY 

Nine-Point  Circle  Theorem 

Th.  2.     If,  in  a  A  ABC,  ^ 

H,  L,  and  T  be  feet  of 
altitudes ;  M,  E,  and  D  feet 
of  medians,  and  S,  O,  and 
F  mid-points  of  BK„j  CK^, 
and  AK„,  respectively, 

Cone. :  then  H,  L,  T,  M,  Ey  D,  S,  Gy  F  are  cyclic. 

Dem.     1.   Pass  a  circle  through  M,  E,  and  D. 
First,   prove  that  this   circle   passes   through   H]    second, 
through!^.  ME  W  b  and  MDW  a. 

[The  mid-join  of  2  sides  of  a  A  is  II  to  the  3d  side.]  (VII.  Ex.  8.) 
.-.  4-side  MEGD  is  a  parallelogram.  (Def.  of  O.) 

.-.  Z  DME  =  Z  ACB.  (VI.  1'  a'.) 

A  DHC  is  isosceles.  (VII.  4.) 

.-.  Z  DHC  =  Z  DCH,  and  similarly,  Z  CHE  =  Z  HCE. 
.-.  ZDHCi-^CHE=Z.DCH+ZHCE{=Z.ACB).    (Ax. 2.) 

.-.  z  z>ir^  =  z  i)io;.  (Ax.  i.) 

.-.  a  circle  through  M,  E,  and  D  passes  through  H. 

[The  arc   DME   is   the    locus    of    vertices   of    all  angles 

=  ZDME,  etc.] 

Similarly,  this  identical  circle  passes  through  L  and  T. 

Q.E.D. 
2.    Draw  the  joins  TF  and  ZD. 

TF  divides  rt.A  TAK„  into  two  isosceles  triangles.   (Why  ?) 

LD  divides  rt.  A  LCA  into  two  isosceles  triangles. 

These  two  triangles  have  Z  CAL  in  common,  and  are  there- 
fore mutually  equiangular. 

.-.  Z  TEA  =  Z  ADL,  whose  supplements,  A  TFL  and  TZ)!., 
are  therefore  equal. 

.-.  a  circle  passing  through  T,  D,  and  L  must  pass  through  F. 

Similarly,  it  may  be  shown  that  this  circle  also  passes  through 

S  and  O, 

Q.E.D. 


XII.     MEASUREMENT  OF  ANGLES  129 

Theorem  of  Orthogonal  Circles 

Th.  3.    If  circles  of  a  given  radius  are  drawn  so  as  to  cut 
a  given  O  Z  orthogonally, 


Cone. :  then  the  locus  of  their  centers  is  a  circle  concentric 
with  Z,  whose  radius  is  the  hypotenuse  of  a  right  triangle 
whose  legs  are  the  radius  of  O  Z  and  the  given  radius. 

Dem.  Draw  any  radius  of  Z,  as  KP,  and  at  P  erect  a  per- 
pendicular to  KP  and  equal  to  the  given  radius. 

With  K  as  a  center  and  KL  as  a  radius,  describe  the  O  M. 

This  circle  is  the  locus  of  the  centers  described  in  the  hy- 
pothesis, for 

XPand  KP  are  of  constant  lengths,  and  the  Z  LPK  is  always 
a  right  angle. 

.-.  LK,  the  hypotenuse,  must  always  be  of  the  same  length. 

.*.  the  O  Jf,  with  the  center  K  and  a  radius  equal  to  LK,  is 

the  locus  required. 

^  Q.E.D. 


Ex.  26.  If  the  angle  formed  by  two 
tangents  is  50°,  how  many  degrees  in  each 
of  the  intercepted  arcs  ? 


Ex.  27.  A  circle  is  circumscribed  about  a  triangle. 
Prove  that  the  radii  drawn  to  the  extremities  of  the 
base  form  an  angle  equal  to  twice  the  angle  at  the 
vertex  of  the  triangle. 


130  THE  ELEMENTS  OF  GEOMETRY 

Classification  op  Problems  —  Indeterminate, 
Determinate,  and  Overdeterminate 

Kinds  of  Equations 

In  algebra  the  student  has  learned  that  there  are  three 
classes  of  simultaneous  equations,  to  wit : 

Indeterminate,  having  an  indeterminate  number  of  roots. 

Determinate,  having  a  determinate  number  of  roots. 

Overdeterminate,  having,  in  general,  no  roots. 

So  in  geometry  problems  may  be  similarly  classified,  to  wit : 

Indeterminate,  in  which  too  few  conditions  are  imposed  to 
give  a  determinate  or  definite  number  of  figures  that  will  satisfy 
the  given  conditions. 

E.g.  draw  a  circle  tangent  to  a  given  line  at  a  given  point  in  the  line. 
An  indeterminate  number  of  such  circles  may  be  drawn. 

Determinate,  in  which  enough  conditions  are  imposed  to  give 
a  determinate  number  of  figures  that  will  satisfy  the  given 
conditions. 

E.g.  draw  a  circle,  with  a  given  radius,  that  shall  be  tangent  to  a  given 
line  at  a  given  point  in  the  line.    Two  such  circles  may  be  drawn. 

Overdeterminate,  in  which  too  many  conditions  are  imposed. 

E.g.  draw  a  circle,  with  a  given  radius,  that  shall  be  tangent  to  a  given 
line  at  a  given  point  in  the  line,  and  at  the  same  time  (simultaneously) 
pass  through  a  given  point.     In  general,  no  such  circle  can  be  drawn. 

EXERCISES  IN  INDETERMINATE,   DETERMINATE,   AND 
OVERDETERMINATE  PROBLEMS 

1.  Add  to  the  following  indeterminate  problems  a  condition  that  will 
make  each  determinate  : 

(a)  Draw  a  line  through  a  given  point. 

(b)  Draw  a  perpendicular  to  a  given  line. 

(c)  Draw  a  circle  tangent  to  two  intersecting  lines. 

(d)  Construct  a  A,  having  given  one  side  and  an  angle  adjacent  to  it. 

(e)  Construct  a  triangle,  having  given  2  A. 

2.  By  what  axiom  is  the  following  problem  determinate  ? 
Draw  the  bisector  of  the  vertex  angle  of  a  given  triangle. 


XII.     MEASUREMENT  OF   ANGLES 


131 


3.  Why  is  the  following  problem  overdeterminate  ?  Draw  a  bisector 
of  the  vertex  angle  of  a  triangle  that  shall  be  perpendicular  to  the  base. 

4.  Arrange  a  summaiy  of  quadrilateral  relations,  similar  to  that  of  tri- 
angular relations  on  page  99,  giving  ten  properties  of  the  angles  and  lines 
of  a  4-side. 

Problems  —  Their  Classification  Illustrated 


Indeterminate 

Determinate 

Overdeterminate 

Construct : 

Construct : 

Construct : 

1. 

A  point  a  distant 

A  point  a  distant 

A  point  a  distant 

from  P. 

from    P,  and    b 

from  P,  b  distant 

distant  from  P^. 

from  Pi,  and  c 
distant  from  P. 

2. 

A  point  a  distant 

A  point  a  distant 

A  point  a  distant 

from     a     given 

from  I,  and  b  dis- 

from I,  b  distant 

line  I. 

tant  from  P. 

from  F,  and  c  dis- 
tant from  Pi. 

3. 

A  point  a  distant 

A  point  a  distant 

A  point  a  distant 

from     a     given 

from  O  K,  and  b 

from  Q  K,b  dis- 

OK. 

distant    from    a 

tant  from  I,  and  c 

given  line  I. 

distant  from  l^. 

4. 

A    point    equidis- 

A   point    equidis- 

A point  equidistant 

tant    from    two 

tant    from    two 

from   two   paral- 

parallels. 

parallels,  and   a 

lels,     a     distant 

.  distant         from 

from  OA"  and  b 

OK 

distant  from  P. 

5. 

A    point    equidis- 

A   point    equidis- 

A point  equidistant 

tant    from    two 

tant    from     two 

from  two  concen- 

concentric     cir- 

concentric     cir- 

tric circles  whose 

cles  whose  radii 

cles  whose  radii 

radii  are  a  and  6, 

are  a  and  b,  re- 

are a  and  b,  and 

and  at  the  same 

spectively. 

at  the  same  time 

time     c     distant 

•  c    distant    from 

from  OK  and  e 

QK 

distant  from  I. 

182  THE  ELEMENTS  OF  GEOMETRY 

Ex.  28  (a).   Parallel  chords  intercept  equal  arcs. 

Ex.  28  (6).    If  the  opposite  ends  of  two  parallel 
chords  are  joined,  two  isosceles  triangles  are  formed. 


Ex.  29.  If  a  4-side  is  circumscribed  about  a  circle,  prove  that  the  sum 
of  two  opposite  sides  equals  the  sum  of  the  other  two  sides. 

Ex.  30.    If  A  is  any  point  in  a  diameter,  B  the    (j 
extremity  of  a  radius  perpendicular  to  the  diameter, 
E  the  point  in  which  AB  meets  the  circumference,  C 
the  point  in  which  the  tangent  through  E  meets  the 
diameter  produced,  then  AC  =  EC. 


Ex.  31.  If  two  circles  are  internally  tangent,  and 
the  diameter  of  the  less  equals  the  radius  of  the  larger, 
the  circumference  of  the  less  bisects  every  chord  of  the 
larger  which  can  be  drawn  through  the  point  of  contact. 


Ex.  32.  Two  circles  are  internally  tangent  in  the 
point  E,  and  AB  is  a  chord  of  the  larger  circle 
tangent  to  the  less  in  the  point  C.  Prove  that  EC 
bisects  Z^i?5. 


Ex.  33.  Show  that  in  a  circumscribed  hexagon  the  sum  of  one  set  of 
alternate  sides  (first,  third,  fifth)  equals  the  sum  of  the  other  set  (second, 
fourth,  sixth). 

Show  also  that  the  sum  of  one  set  of  alternate  sides  of  a  circumscribed 
octagon  equals  the  sum  of  the  other  set. 

Ex.  34.  Show  that  any  circumscribed  polygon  with  an  even  number 
of  sides  has  the  sum  of  one  set  of  alternate  sides  equal  to  the  sum  of  the 
other  set. 

Ex.  35.  Inscribe  a  square  in  a  given  circle.  Show  how  to  obtain  from 
this  square,  by  bisection  of  sides,  etc.,  a  regular  inscribed  octagon  and  a 
regular  inscribed  hexa-decagon. 


Ex.  36.   An  equilateral  inscribed  polygon  is  equiangular. 


XII.     MEASUREMENT  OF  ANGLES  183 

Ex.  37.  What  is  the  converse  of  Ex.  36  ?  Is  it  true  ?  Illustrate  your 
answer  by  a  figure. 

Ex.  38.  If  through  one  of  the  points  of  intersec- 
tion of  two  circles  tlie  diameters  of  the  circles  be 
drawn,  the  join  of  the  other  extremities  of  these 
diameters  passes  through  the  other  point  of  intersec- 
tion of  the  circles. 

Ex.  39.  The  join  spoken  of  in  Ex.  38  is  parallel  to  the  line  of  centers 
of  the  circles. 

Ex.  40.  This  join  is  longer  than  any  other  line  through  a  point  of 
intersection  of  the  circumferences  and  terminated  by  them. 

Ex.  41.    What  is  a  cyclic  4-side  ? 
■  Ex.  42.    What  kind  of  angle  with  reference  to  the  circle  is  AECB'i 
ZECF2 

Ex.  43.    What  is  the  measure  of  Z  ECF  ?     Why  ? 

Ex.  44.    What  is  the  measure  of  Z  EFC  ? 

Ex.  45.    What  is  the  measure  of  Z  00^  ?     Oi  ZEQO? 

Ex.  46.    If    FH  bisects  ZF  and    ML 
'bisects  Z  M,  show  that : 
(a)  Arc  LE    —  arc  CJ=  arc  LA  —  arc  JB. 
(6)  Arc  LE    +  arc  JB  =  arc  LA  +  arc  CJ. 
(c)  Arc  LEC  +  arc  JB  =  arc  LA  +  arc  EC  J. 

Ex.  47.    The  first  member  of  (c)  is  the 
measure  of  what  angle  ? 

Ex.  48.    The  second  member  of  (c)  is 
the  measure  of  what  angle  ? 

Ex.  49.    Therefore,  what  kind  of  triangle  is  FOQ  ? 

Ex.  50.   Why,  then,  is  FH  ±  LM? 

Ex.  51.    Similarly,  prove  that  A  MGK  is  isosceles. 

Ex.  52.    Combine  41,  46,  and  50  into  one  theorem. 

Ex.  53.  If  two  straight  lines  are  drawn  through  the  point  of  contact  of 
two  tangent  circles,  the  chords  of  the  arcs  intercepted  by  these  lines  are 
parallel. 

Ex.  54.    What  parallelograms  may  be  inscribed  in  a  circle  ? 

Ex.  55.  The  apparent  size  of  a  circular  object  is  determined  by  the 
angle  between  two  tangents  drawn  from  the  eye  to  the  object. 

What  is  the  locus  of  the  point  from  which  a  given  circle  always  appears 
to  have  the  same  size  ? 


134 


THE  ELEMENTS  OF  GEOMETRY 


B 


Ex.  56.   Given  the  rt.  A  ABC. 

At  any  point  //  of  the  hypotenuse  AB  erect  a 
±HE. 

Let  HE  intersect  BC  (produced)  in  F. 

Draw  AF  and  GB. 

Let  OB  (produced)  meet  AF  in  Q. 

As  the  ±HE  moves  along  AB,  what  is  the  locus 
oiQ?  AM 

Ex.  67.   What  part  of  the  hypotenuse  of  a  right  triangle  is  the  median 
to  the  hypotenuse  ? 

Ex.  58.   Into  what  kind  of  triangles  does  the  median  to  the  hypotenuse 
divide  the  right  triangle  ? 

Ex.  59.   Draw  a  right  triangle  and  its  altitude  to  the  hypotenuse.   Name 
the  three  sets  of  complemental  angles  in  your  figure. 

Ex.  60.    What  is  the  measure  of  an  inscribed  angle  ? 

If,  in  the  adjacent  figure,  CE±  BA,  OF  A.  BE,  and 
the  4-side  is  inscribed,  point  out  three  angles  equal  to 
Z  BOF^  and  give  reasons. 

Ex.  61.    Give  two  reasons  why  ^FEO  =  Z  CAO. 

Ex.  62.   Prove  that  if  the  diagonals  of  a  cyclic 
4-side  be  perpendicular  to  each  other,  and  from  their 
intersection  a  perpendicular  be  let  fall  on  one  side  of  the  4-side,  this  per- 
pendicular will  bisect  the  opposite  side. 

In  the  following  exercises  the  angles,  sides,  and  principal  lines  of  a 
triangle  are  represented  thus  : 

Angles:  A,  B,  C. 

Sides :  a,  b,  c  opposite  ZA,  ZB,  and  Z  C,  respectively. 

Altitudes :  ha,  hb,  and  he,  altitudes  to  sides  a,  b,  and  c,  respectively. 

Medians :  nia,  W5,  and  »ic,  medians  to  sides  a,  b,  and  c,  respectively. 

Bisectors  of  A :  t^,  ta,  and  ta^  bisectors  of  A,  B,  and  C,  respectively. 

(Note.  —  When  one  angle  of  an  isosceles  triangle  is  given,  all  the  angles 
are  given.) 

Construct  an  isosceles  triangle,  given : 

Ex.  63.   c  and  ZA.  Ex.  64.    c  and  Z  C. 

Ex.  65.   c  and  the  radius  of  the  inscribed  circle. 

Ex.  66.   c-\-  a  and  Z  B. 
(Anal. :  Extend  c  to  C,  making  BC  =  c -\- a.    Draw  CC.    Use  IIL  2  o») 

Ex.  67.   a  and  he.  Ex.  69.    Z  B  and  he. 

Ex.  68.   c  and  he.  Ex.  70.   Z  C  and  a  4-  6. 


',^^^^ 


XII.     MEASUREMENT   OF  ANGLES  135 

Ex.  71.   c  and  hi.  Ex.  73.   Z  C  and  a  +  c. 

Ex.  72.   he  and  Z  C  Ex.  74.   Z  C  and  wij. 

Ex.  75.    6  +  /ic  and  Z  C. 
(Anal.  :  Let  /ic  with  its  extension  to  ^=^c+&-    Draw  WA.    Use  III.  2  a.) 

Ex.  76.    6  +  Ac-  and  c. 
/Anal.  :  Let  Ac  with  its  extension  to  JBT  =  Ac  +  6.     Take  -•    Use  III.  2.  j 

Ex.  77.   a  4-  6  4-  c  and  Z  A. 

(Note.  —  If  one  acute  angle  of  a  right  triangle  is  given,  the  other  is 
also  given.) 

Construct  a  right  triangle  (right  angle  at  C),  given: 

Ex.  78.   c  and  Z  A.  Ex.  79.   c  and  Ac. 

Ex.  80.   c  and  the  radius  of  the  inscribed  circle. 

Ex.  81.   Z  A  and  the  radius  of  the  inscribed  circle. 

Ex.  82.    a  and  the  radius  of  the  inscribed  circle. 

Ex.  83.  rtic  and  Ac.  (Anal. :  Use  VIL  4.) 

Ex.  84.   Z^and  Ac. 

Ex.  85.   The  two  segments  of  c  mad6  by  Ac. 

Ex.  86.    The  two  segments  of  c  made  by  tc. 

(Anal.  :  Extend  tc  to  meet  O  on  c.     Use  XII.  2.) 
Ex.  87.   c  and  the  distance  from  the  vertex  of  Z  C  to  a  given  line. 
Ex.  88.    a  +  h  and  Z  A. 

Ex.  89.  The  radius  of  the  inscribed  and  the  radius  of  the  circumscribed 
circle. 

Ex.  90.   a  —  h  and  c. 
Ex.  91.    c  —  a  and  Z  A.    • 

(Anal. :  Z  5  is  known.    Const,  a  A,  given  c  —  a  and  its  A.) 
Ex.  92.   a  +  6  +  c  and  AA. 
Construct  an  equilateral  triangle,  given : 

Ex.  93.   The  perimeter.  Ex.  94."  The  altitude.         Ex.  95.    a  +  A. 

Ex.  96.   The  radius  of  the  inscribed  circle. 
Ex.  97.   The  radius  of  the  circumscribed  circle. 
Construct  a  triangle,  given : 
Ex.  98.   The  perimeter  and  Z  A  and  Z  B. 
Ex.  99.   c,  Ac,  and  Z  C.  Ex.  100.   c,  nic,  and  Z  C. 

Ex.  101.   c,  Ac,  and  ha.  (Use  IX.  4.) 


136  THE   ELEMENTS  OF  «^;E0]VIETRY 

Ex.  102.  ./ie,  ^Ay  and  Z  B.  Ex.  104.   a,  c,  and  mo. 

Ex.  103.    a,  hei  and  c.  Ex.  105.    c,  Ac»  and  wip. 

Ex.  106.    a,  6,  and  Wc.        (Double  the  median.     Use  Conv.  of  VI.  3.) 

Ex.  107.   »/io  /Ja»  and  /ij.  (Double  the  median.    Use  IX.  4.) 

Construct  a  square,  given : 

Ex.  108.    Its  apothem. 

Ex.  109.   The  difference  between  its  diagonal  and  its  side. 

Construct  a  rhombus,  given  : 

Ex,  110.   The  two  diagonals.  Ex.  111.  One  diagonal  and  a  side. 

Construct  a  trapezoid,  given :  -ft/^5___£ — --::^np 


.^^^- 


Notation :  ^^^^""""'^''^^^^^^ 

Ex.  112.    a,  c,  ZA,  and  ZB.  A                  '^              B 

Ex.113,    a,  c,  A,  and  Z  A  Ex.115,    a,  6,  c,  and  ^^. 

Ex.  114.    c,  e,  AC,  and  BE.  Ex.  116.   c,  a,  &,  e,  and  AC. 
Construct  a  rhomboid,  given : 

Ex.  117.    c  and  ^C  and  BE.  Ex.  119.    AC,  h,  and  ZA. 

Ex.  118.    c,  AC,  and  /i.  Ex.  120.   AC,  h,  and  b. 

Ex.  121.   If  a  4-side  is  circumscribed  about  a  circle,  the  central  angles 
subtended  by  the  opposite  sides  are  supplemental. 

Ex.  122.    In  rt.  A  ABC,  inscribe  OK. 

Ex.  123.    If  the  points  of  tangency  of  this  O  be  N,  G,  and  T,  why 
does  BG  =  BN? 

Ex.  124.   Show  that  4-side  KGATia  a  square. 

Ex.  125.    Show  CB  =  CA  +  BA  minus  the  diameter  of  the  inscribed 
circle. 

Ex.  126.    Show  that  the  diameter  of  the  circumcircle  plus  the  diameter 
of  the  incircle  equals  the  sum  of  the  legs  of  the 
right  triangle. 

Ex.  127.  If  AB  is  produced  to  the  left,  mak- 
ing AC'  =  AC,  what  is  the  value  of  Z  AC  C? 

Ex.  128.  If  BC  is  the  sum  of  the  legs  in  the 
Tt.AABC,  and  BC  is  the  diameter  of  the  cir- 
cumcircle, what  are  the  two  loci  of  C  ? 

Ex.  129.    Construct  a  right  triangle,  given  the  sum  of  the  legs  and  the 
radius  of  the  incircle. 


XII.     MEASUREMENT  OF  ANGLES 


137 


Ex.  130.  If  the  radius  of  the  circum-O  is  given,  and  also  that  of  the  in-O, 
how  from  these  do  you  obtain  the  hypotenuse  and  the  sum  of  the  legs  ? 

Ex.  131.  Construct  a  right  triangle,  given  the  radius  of  the  incircle 
and  radius  of  the  circumcircle. 

Ex.  132.  If,  in  A  ABC,  CT  bisects  ZACB  and  AL  is  drawn  perpen- 
dicular to  or,  by  what  theorem  is  A  ^C^  isosceles  ? 

Ex.  133.    If  M  is  the  mid-point  of  AB,  why  is  LM  parallel  to  QB  ? 

Ex.  134.   Prove  that  LM  =  ^(a-b). 

Ex.  135.  If  BL'  is  drawn  perpendicular  to  CT,  prove  that  the  join 
L'M  is  also  =  J(a  —  b). 

Ex.  136.  If  CT  bisect  ext.  Z  BCA',  and  from 
J5  a  ±  to  CT'  is  drawn,  then  the  join  of  M  and 
the  foot  of  this  perpendicular  =  l(a  -\-h). 

Ex.137.    ZLAM=ZCAB-ZCAQ. 
then,  that      z LAM  =  \{Z  CAB  -  ZB). 

Ex.  138.    If  CH±AB,  prove  that 

ZHCT=^X^C^S-^S)- 

Ex.  139.    State  the  preceding  theorem  in  general  terms. 

Ex.  140.   Draw  FQ,  and  prove  that  as  AAFQ  is  isosceles  (Why  ?), 
ZQFB  =  ZCAB-ZB. 

Ex.  141.    Show  that  Z  AQB  =  Z  ACB  -{-  ^  the  supplement  of  ZACB. 

Ex.  142.  Construct  a  triangle,  having  given  the  base,  the  difference  of 
the  two  sides,  and  the  vertex  angle. 

Ex.  143.  If  from  Jf,  the  middle  point  of  the  arc  AMB,  any  two  chords 
MF,  MQ  are  drawn,  cutting  the  chord  AB  in  E  and  0,  then  CEFG  is 
cyclic. 


XIII.  GROUP  ON  AREAS  OP  RECTANGLES 
AND  OTHER  POLYGONS 

(Briefly,  the  Areal  Group) 

DBTINITIONS 

The  Area  of  a  plane  figure  is  the  ratio  of  its  surface  to  some 
other  surface  taken  as  a  unit  of  area. 

This  Unit  of  Area  is  usually  a  square  whose  base  and  altitude 
are  each  a  unit  of  length. 

A  Polygon  is  circumscribed  to  a  circle  when  the  sides  of  the 
polygon  are  tangents  to  the  circle. 

A  Polygon  is  inscribed  in  a  circle  when  the  sides  of  the 
polygon  are  chords  of  the  circle,  that  is,  when  its  vertices  are 
in  the  circumference. 

PROPOSITIONS 

XIII.  1.   Two  rectangles 

(a)  having  equal  bases,  vary  as  their  altitudes  ; 
(6)  having  equal  altitudes,  vary  as  their  bases. 


Hyp.L  (a)  If, 
in  the  a's  A-Q 
and  E-L,  the 
bases  AB  and  EF 
are  equal, 


M 


IT 


A  BE  M 

Cone. :  then  □  A-G :  czi  E-L : :  altitude  AM:  altitude  EH. 

138 


^^>^ 


XIII.     AREAS  OF   POLYGONS 


139 


Case  I.     Commensurable  case. 

Dem.     If  AM  and  EH  have  common  measure  (v.  XI.  Prob.  I.), 
say  m,  apply  it  as  many  times  as  possible  to  AM  and  to  EH. 
Suppose  it  is  contained  in  AM  seven  times,  in  EH  five. 

r.  AM:EH::7:5. 

Through  the  points  of  division  draw  parallels  to  the  bases. 
The  cjA-C  will  be  divided  into  seven  ^  n's,  and  the  nuE-L 
will  be  divided  into  five  ^  a's.  (VI.  4  a.) 

Furthermore,  all  of  these  smaller  a's  are  ^.  (VI.  4  a.) 

.-.  rDA-C:u3E-L::7:5. 

.•    □  A-C:\I3E-L  : ;  altitude  AM:  altitude  EH.     (Ax.  1.) 

Q.E.D. 

K  ■      E 


G  A 

Case  II.     Incommensurable  case ;  that  is,  h  and  hi  have  no 
common  measure. 

Dem.     Divide  AB  into  any  number  of  equal  parts,  say  n. 
Apply  one  of  these  as  a  divisor  to  CE  until  there  is  a  re- 
mainder LE  less  than  the  divisor. 
Dmw  LMWHC. 

□  G-B :  n  0-M: :  AB :  CL.     (XIII.  1.  (a)  Case  I.) 

Now,  if  we  decrease  the  divisor,  we  decrease  the  remainder 
without  affecting  the  equality  of  the  quotients. 

That  is,  as  LE  =  0. 

and  CL  =  CE, 

so  cnCM=u3  CK.     (XI.  Def .  of  Limit.) 


140 


THE  ELEMENTS  OF  GEOMETRY 


The  ratios  □  G-B  :  □  C-M  and  AB :  GL,  however,  remain 
equal  as  they  approach  their  limits,  viz. : 


□  G-B :  □  C-Zrand  AB :  CE. 
'.  UDG-B:n2C-K::AB:CE. 


(Case  I.) 


[If,  while  approaching  their  respective  limits,  etc.] 

(XI.  Post.  Limits.)     Q.E.D. 
Hyp.  1.    (b)  If,  I  H 

in  a's  I  and  II, 
the  altitudes  h,  h^ 
are  equal,  and  the 
bases  are  b  and  61, 


/( 


Cone. :  then 


□  I:nII::6:6i. 


Dem.     The  proof  of  1.  (b)  is  exactly  similar  to  that  of  1.  (a). 

Q.E.D. 

XIII.  1  a.  Any  tivo  rectangles  are  to  each  other  as 
the  products  of  their  bases  and  altitudes. 


Hyp.  If  the 
two  a's  I  and  II 
have  bases  b  and 
61,  and  altitudes 
h  and  ^1, 

Cone:  then 


/ 

/', 

// 

//, 

III 

b  b, 

unl'.nnlli'.b  'h:,bi'\. 


Dem.     Construct  a  □  III  with  base  equal  to  b  and  altitude 

equal  to  ^1. 

□  I :  □  III :  :  7i :  ^1.     (1)       (XIII.  1  (a).) 

□  III :  □  II :  :  6  :  6,.       (2)       (XIII.  1  (6).) 
Multiply  proportion  (1)  by  (2),  and  we  have 
□  I :  □  II : :  6  •  7i :  61  •  ^1. 


[If  two  proportions  be  multiplied  together,  etc.] 


(XI.  3.) 

Q.E.D. 


XIII.      AREAS   OF   POLYGONS  141 

XIII.  1  h.  Any  two  parallelograms  are  to  each  other 
as  the  products  of  their  bases  and  altitudes. 


Hyp.  If  two  UJA-B  and  C-E  have  b  and  b^,  h  and  Ji^  as 
bases  and  altitudes,  respectively, 

Cone. :  then     O  A-B  :  EJ  C-E  -.'.b-h-.b^-  hy. 

Dem.  Draw  the  perpendiculars  to  meet  sides  (produced  if 
necessary)  as  in  figure. 

Et.  A  I  ^  rt.  A  I'  and  rt.  A II  ^  rt.  A II'. 

[If  two  rt.  A  have  the  leg  and  hypotenuse  of  one,  etc.]  (V.  4.) 

.-.  r]A-B=^\ziA-R)  CJC-E  =  C3C-H. 

b  and  7i  are  identical  in  O  A-B  and  □  A-R. 
bi  and  h^  are  identical  in  O  C-E  and  □  C-H. 

Now  □  A-R  :  □  C-H  ::b'Ji:b,'  h,.      (XIII.  1  a.) 

.'.  CJA-B  :  O  C-E  ::b'h:b,'  K 

Q.E.D. 

ScH.     A  parallelogram  is  equal  to  a  rectangle  of  the  same 

base  and  altitude. 

Ex.  1.  Two  rectangles  are  equal.  The  bases  are  27  and  18,  respec- 
tively, and  the  altitude  of  the  first  is  8.  What  is  the  altitude  of  the  second  ? 

Ex.  2.  Construct  three  equal  triangles  on  the  same  base,  the  first  of 
which  shall  be  acute,  the  second  right,  and  the  third  obtuse. 

Ex.  3.   A  number  of  equal  triangles  stand  on  the  same  base. 

How  do  their  altitudes  compare  ? 

Show,  then,  that  the  locus  of  their  vertices  consists  of  two  lines  parallel 
to  the  base. 

Ex.  4.  The  base  of  a  triangle  is  divided  into  five  equal  parts,  and  the 
points  of  division  are  joined  to  the  vertex. 

How  do  the  altitudes  of  the  resulting  triangles  compare  ? 

How,  then,  do  the  areas  of  these  triangles  compare  ? 


142  THE  ELEMENTS  OF  GEOMETRY 

XIII.  1  c.  Any  two  triangles  are  to  each  other  as  the 
products  of  their  bases  and  altitudes. 

Hyp.    lithe  A  ABO      q     .^ 

andEFHhdiveABsiud      /^V  /  ^^(T^^T- n^ 

EF  for  bases,  and  CQ   /  X^/  y'^^'^^^X^ 

and  HA  for  altitudes,    a     Q  B  A    E  F 

Cone:  then  A  ABG:  A  EHF: :  □  oiAB  •  CQ  :  aof  EF  -  HA. 

Dem.     Complete  the  parallelograms  as  in  the  figures. 

OA-MiOE-L::  a  oi  AB  -  CQ:C2  of  EF:  HA. 

(XIII.  1  5.) 

But      A  ABC  is  i  O  A-M  and  A  EFH  is  ^  O  E-L. 

[The  diagonal  of  a  O  divides  it  into  two  ^  A.]   (VI.  1  a,  Sch.) 

.-.  AABO.AEHF:  .nnoiAB-  CQ:  □  of  EF  -  HA. 

Q.E.D. 

Sch.  1.  Parallelograms  (or  triangles)  with  equal  bases  are 
to  each  other  as  their  altitudes. 

Sch.  2.  Parallelograms  (or  triangles)  with  equal  altitudes 
are  to  each  other  as  their  bases. 

Sch.  3.  If  parallelograms  (or  triangles)  have  equal  altitudes 
and  equal  bases,  they  are  equal. 

Ex.  5.  How  would  you  divide  a  triangle  into  n  equal  parts  by  lines 
passing  through  the  vertex  ? 

Ex.  6.  If  the  altitude,  26  ft.,  of  a  rectangle  is  to  be  reduced  to  20  ft, 
how  much  must  be  added  to  the  base,  30  ft.,  to  keep  the  area  un- 
changed ? 

Ex.  7.  If  one  angle  of  a  right  triangle  be  30°,  how  does  the  hypotenuse 
compare  with  the  shortest  side  ? 

Find  the  area  of  a  right  triangle,  one  of  whose  sides  is  12  ft.  and  one 
of  whose  angles  is  30°. 

Ex.  8.  Show,  by  drawing  a  figure,  that  the  square  on  one  half  a  line  is 
one  fourth  the  square  on  the  line. 

Ex.  9.  Show  also  that  the  square  on  one  third  a  line  is  one  ninth  the 
square  on  the  line. 


XIII.     AREAS   OF   POLYGONS 


143 


XIII.  2.   Tlie   area   of  a  parallelogram   equals    the 
product  of  its  base  and  altitude. 

Hyp.    If  O  A-0  has  ^ 

6  as  a  base  and  h  as  an 
altitude, 


r      1  \ 


Cone. :  then  the  area  of  O  A-G  —h'Ji. 

Dem.     On  any  line  as  UN  assumed  as  a  unit  of  length  (v.  Def. 
in  XI.)  construct  2,UU-T. 

This  square  may  be  taken  as  the  unit  of  area  (v.  Def.  in  XIII.). 

.-.  the  n  U-T=l. 

nJA-CiU  U-T::b-h:l  x  1. 

[Any  two  UJ  are  to  each  other,  etc.]  (XIII.  1  b.) 

.'.  O  A-0 : 1 : :  6  •  7i :  1 ;  that  is,  the  area  of  O  A-C  =  b  -h. 

(Def.  of  area  (XIII.).) 
Q.E.D. 

XIII.  2  a.   The  area  of  a  triangle  equals  one  half  the 
product  of  its  base  and  altitude. 

Hyp.     If  b  and  h  are 

the   base   and  altitude, 
respectively,  of  A  ABC, 

Cone. :  then  the  area  of  A  ABO  =^b  -h. 
Dem.     Complete  UJ  A-E  as  in  figure. 

AABO  =  ^IZIA-E. 

[The  diagonal  of  a  O  divides  it  into  2  ^  A.]      (VI.  1,  Sch.) 
The  area  of  O  A-E  =  b'h.  (XIII.  2.) 


.-.  the  area  of  A  ABO=  \b'K 


Q.E.D. 


Ex.  10.  The  legs  of  one  right  triangle  are  8  ft.  and  6  ft. ;  of  another  5  ft. 
and  12  ft.     What  is  the  ratio  of  their  areas  ? 

Ex.  11.  A  parallelogram  has  a  base  of  9  ft.  and  an  altitude  of  16  ft. 
What  is  the  side  of  a  square  equivalent  to  the  parallelogram  ? 


/ 


144  THE  ELEMENTS  OF  GEOMETRY 

Areas  of  Irregular  Figures 

ScH.  To  obtain  the  area  of  an  irregular  polygon,  join  any 
point  within  the  polygon  with  the  vertices  of  the  figure. 

Find  the  area  of  each  triangle  thus  formed.  The  sum  of 
these  areas  equals  the  area  of  the  polygon. 

XIII.  2  h.  Tlie  area  of  a  trapezoid  equals  the  product 
of  the  altitude  and  the  mid-join  of  the  non-parallel  sides. 

Hyp.    If  the 
4-side  A-C  is  a  -9 _-**— 

trapezoid,  MJ 
its  mid-join, 
and  Ci^  its  alti- 
tude. 

Cone. :  then  the  area  of  trapezoid  A-C  =  □  of  CF  •  MJ, 

Dem.     Draw  AC. 

The  area  of  A  ACB  =CF'^'  (XIII.  2  a.) 

The  area  of  A  AEQ  =.CF'^'  (XIII.  2  a.) 

.-.  A^OB  +  AACE  =CF-^^'^  ^^.  (Ax.  2.) 

That  is,  the  area  of  trapezoid  A-C  =  OF-  ^^  +  C^ 
But  4B±CE^^j 

[The  mid-join  of  a  trapezoid  =  \  the  sum  of  II  sides.]    (VII.  3  &.) 

.-.  the  area  of   the  trapezoid  A-G  equals  the  rectangle  of 

CF'MJ. 

Q.E.D. 

Ex.  12.  In  the  last  problem  substitute  "triangle"  for  "parallelo- 
gram," and  solve. 

Ex.  13.  If  the  area  of  a  trapezoid  be  65  sq.  ft.,  and  the  parallel  sides 
respectively  10  ft.  and  16  ft.,  vs^hat  is  the  altitude  ? 

Ex.  14.  A  square  and  a  rectangle  have  the  same  perimeter.  Which 
hais  the  greater  area  ?    Why  ? 


XIII.     AREAS  OF  POLYGONS 


145 


XIII.  3.  The  area  of  a  circumscribed  polygon  equals 
one  half  the  ijroduct  of  its  perimeter  and  the  radius  of 
the  inscribed  circle. 

Hyp.  If  an  n-gon 
ABC-F  is  circum- 
scribed to  a  circle 
of  center  7ij  and  of 
radius  r^, 

Cone. :  then  the  area  of  n-gon 

ABC-F  =  i(AB  ■^BC+CE+  "')r,. 
Dem.     Draw  K^A,  KiB,  etc. ;  also  K^L,  KtM,  etc. 
K^L,  KiM,  etc.,  are  altitudes  of  AAK^B,  BK^C,  etc.     (IX.  4.) 
That  is,  the  radius  of  the  inscribed  circle  equals  the  altitude 
of  the  triangles  that  make  up  the  n-gon. 

But  the  area  of  AAKiB  =  —'^\. 

Similarly,  for  the  remaining  triangles. 

.-.  the  sum  of  the  areas  of  the  triangles,  or  the  area  of  the 

n-gon  ABC-F  =  i {AB  +  BC  +  CE  +  "-)  -  r^. 

Q.E.D. 

Ex.  15.  What  is  the  area  of  a  triangle  if  the  base  is  1384  ft.,  and  the 
altitude  is  256  ft.  ? 

Ex.  16.   What  is  the  area  of  a  rt.  isosceles  A  one  leg  of  which  is  1414  ft.  ? 

Ex.  17.  What  is  the  area  of  a  right  triangle  whose  perimeter  is  840  ft., 
and  whose  sides  are  to  each  other  as  3  :  4  :  5  ? 

Ex.  18.  What  is  the  area  of  a  right  triangle  whose  perimeter  is  300  ft., 
if  its  sides  are  as  5  :  12  :  13  ? 

Ex.  19.  The  base  of  a  triangle  is  to  its  altitude  as  11  :  :  60  ;  the  area  of 
the  triangle  is  1320  sq.  ft.     What  is  the  length  of  the  base  ? 

Ex.  20.  The  altitude  of  a  trapezoid  is  16  ft.  ;  the  mid-join  is  32  ft. 
What  is  the  area  of  the  trapezoid  ? 

Ex.  21.  A  A  whose  altitude  is  10  ft.  and  base  24  ft.  is  transformed  into  a 
rhombus.    Its  longer  diagonal  is  16  ft.    What  is  the  length  of  the  shorter? 

What  are  the  dimensions  of  a  rectangle  if  the 

Ex.  22.   Area  is  3822  sq.  ft.  and  the  sides  are  as  6  :  13  ? 

Ex.  23.    Area  is  59100  sq.  ft.  and  perimeter  994  ft.  ? 


146 


THE  ELEMENTS  OF  GEOMETRY 


b 

c        c 

1) 

h     h 


Ex.  24.  In  a  parallelogram  a  line  is  drawn  that  cuts  off  one  fourth  of 
one  side  and  three  fifths  of  tlie  opposite  side.  What  part  of  the  parallelo- 
gram is  each  trapezoid  thus  formed  ? 

Ex.  25.    Show  geometrically  that  if  h  and  c  rep-     c 
resent  lines, 

(6  +  c)a  =  62  +2  6c  +  c\ 

State  the  proposition  in  words,  i.e.  without  the  use 
of  symbols. 


Ex.  26.   Show  geometrically  that  if  6  and  c 
represent  lines, 

(6  -  c)2  =  62  +  c2  -  2  6c.  J 

=  62  _  2  6c  +  c2. 
State  the  proposition  in  words. 


b 

be                 c 

b-c 

c 

(b-c)^ 

bo 

c 

Ex.  27.    Show,  from  the  figure,  that  if  6  and 
c  be  any  two  lines, 

(6  +  c)(6-c)  =  62~c2. 
State  the  proposition  in  words. 


b< 


c        ^ 
I       c     (T 

I  b-c 


b+c 


Ex,  28.  If,  in  a  trapezoid  ABCE^  AC  and  BE^  the  two  diagonals,  are 
drawn,  prove  AABC  =  AABE. 

Ex.  29.  If,  in  the  same  trapezoid,  the  diagonals  intersect  in  M,  prove 
thsii /\A3IE=  A  BMC. 

Ex.  30.  If  two  equal  triangles  have  the  same  base,  and  lie  on  opposite 
sides  of  this  base,  the  join  of  the  vertices  of  the  triangles  is  bisected  by 
the  common  base. 

Ex.  31.  If,  in  a  O  ABCE,  perpendiculars  from  B  and  E  are  let  fall  on 
the  diagonal  AC,  prove  that  they  are  equal. 

Ex.  32.  In  the  above  parallelogram,  if  P  is  any  point  in  the  diagonal 
AC,  prove  that  A  APB  =  A  APE. 

Ex.  33.  Prove  that  if  from  a  vertex  of  a  parallelogram  a  line  is  drawn 
to  the  mid-point  of  one  of  the  opposite  sides,  it  cuts  off  one  third  of  the 
diagonal  it  intersects. 

Ex.  34.  Prove  that  the  line  referred  to  in  the  preceding  theorem  cutg 
off  a  triangle  equal  to  one  fourth  of  the  parallelogram. 


XIII.     AREAS  OF  POLYGONS  147 

XIII.     SUMMARY   OF  PROPOSITIONS  IN   THE  AREAL 

GROUP 

1.  Tloo  rectangles 

(a)  having  equal  bases,  vary  as  their  altitudes, 

(b)  having  equal  altitudes,  vary  as  their  bases. 

a.  Any  two  rectangles  are  to  each  other  as  the 

products  of  their  bases  and  altitudes. 

b.  Any  tivo  parallelograms  are  to  each  other  as 

the  products  of  their  bases  and  altitudes. 

ScH.  A  parallelogram  is  equal  to  a  rectangle  of  the  same 
base  and  altitude. 

c.  Any  two  triangles  are  to  each  other  as  the 

products  of  their  bases  and  altitudes. 

ScH.  1.     Parallelograms  (or  triangles)  with  equal  bases  are 
to  each  other  as  their*  altitudes. 

2.  Parallelograms  (or  triangles)  with  equal  altitudes 

are  to  each  other  as  their  bases. 

3.  Parallelograms  (or  triangles)  with  equal  bases  and 

equal  altitudes  are  equal. 

2.  The  area  of  a  parallelogram  equals  the  product 
of  its  base  and  altitude. 

a    The  area  of  triangle  equals  one  half  the  product 
of  its  base  and  altitude. 
ScH.     Areas  of  irregular  figures. 

b  The  area  of  a  trapezoid  equals  the  product  of 
the  altitude  and  the  mid-join  of  the  non- 
parallel  sides. 

3.  The  area  of  a  circumscribed  polygon  equals  one 
half  the  product  of  its  perimeter  and  the  radius  of  the 
inscribed  circle. 


X   A 


148         THE  ELEMENTS  OF  GEOMETRY 
PROBLEM 

Prob.  I.     To    reduce   a  polygon    to    an    equivalent 

triangle. 

Given.  The  poly- 
gon ABCEF. 

Required.  To  re- 
duce it  to  an  equiva- 
lent triangle. 

Const.     Extend  AF  indefinitely. 

Draw  FC,  a  diagonal. 

Draw  EQ  II  CFj  meeting  AF  produced  in  Q, 

Draw  CQ.  ^  q^^  ^  ^  q^q 

[A  having = bases  and = altitudes  are  =  .]   (XIII.  1.  c,  Sch.  3.) 
.-.  ABCF  +  A  CEF  =  ABCF  +  A  CFQ.        (Ax.  2.) 
.-.  n-gon  ABCQ  =  n-gon  ABCEF. 

Similarly,  draw  OA,  BK,  and  CK. 

The  number  of  vertices  of  n-gon  is  thus  reduced  to  three. 
.-.  A  KCQ  =  n-gon  ABCEF. 

Q.E.P. 

Ex.  35.  What  is  the  centroid  of  a  triangle  ?  (X. ,  6. ) 
If  G  be  the  centroid  of  ABC,  what  is  the  ratio  of  the 
areas  of  ACM  and  AGM?  Of  the  areas  of  BCM  and 
BGM?    Of  the  areas  of  ^  G^^  and  A  CB  ? 

Ex.  36.  How,  then,  do  the  AAGB,  AGC,  and  BGO 
compare  in  area  ? 

State  as  a  theorem  the  property  of  the  centroid  thus  established. 

Ex.  37.  The  centroid  of  a  material  triangle  uniform  in  thickness  and  of 
the  same  material  throughout  is  called  the  "center  of  gravity"  of  the  tri- 
angle.   How  might  this  name  be  suggested  by  the  property  just  established  ? 

Ex,  38.  On  a  side  of  a  given  triangle  as  a  base,  to  construct  a  triangle 
equal  to  the  first  and  having  its  vertex  on  a  given  line. 

Ex.  39.  Show  that  the  greatest  number  of  solutions  possible  in  Ex.  38, 
is  two.  Under  what  conditions  is  the  solution  impossible  ?  Under  what 
conditions  is  the  problem  indeterminate  ?  Draw  figures  to  illustrate  your 
answers. 


XIV.     PYTHAGOREAN   GROUP 


DEFINITIONS 


Projection  on  a  Line 


The  Projection  of  a  Point  on  a  straight  line  is  the  foot  of  the 
perpendicular  from  the  point  to  the  line. 

The  line  on  which  the  perpendicular  is  dropped  is  called  the 
Base  of  Projection. 

The  Projection  of  a  Line-Segment  is  that  portion  of  the  base  of 
projection  which  lies  between  the  projections  of  the  extremities 
of  the  given  line-segment.  _^  2? 


Projection  of  AB 


Note. — This  group  is  named  after  Pythagoras,  an  eminent  Greek 
mathematician  of  the  sixth  century  b.c,  who  was  the  first  to  publish  a 
proof  of  the  first  theorem  of  the  group.  The  truth  of  this  proposition  was 
known  before  his  time,  but  a  proof  had  long  been  sought  in  vain.  The 
theorem,  the  47th  of  Euclid,  is  often  called  the  pons  asinorum  of  geometry. 


Ex.  1.    In  the  right,  acute,  and  obtuse  A  I,  II,  III,  draw  the  projec- 
tions of  a  on  b,  and  of  &  on  a ;  also  of  a  on  c,  and  of  c  on  a. 

C 


140 


150 


THE   ELEMENTS  OF  GEOMETBY 


A 

k     ^^' 

in 

\vr 

l^hv:^.- 

/ 

/ 
/ 

\ 

/ 

\ 

/ 

/ 

\ 

/ 

\ 

1 

\ 

PROPOSITIONS 

XIV.   1.  If  a  triangle  is  right,  the  square  on  the  hy- 
Xiotenuse  equals  the  sum  of  the  squares  on  the  legs. 


Hyp.    If, 

in  a  rt 
AABC,  ZO 
is  the  right 
angle, 


Cone. :  then  the  D  on  c  =  the  D  on  a  4-  the  D  on  h. 

Dem.     Draw  the  altitude  CJ  and  extend  it  to  L  on  the  D  on  c. 

Draw  CHy  CQ,  BE,  and  AF. 

AB  =  AH  and  AE  =  AC.         (Def .  of  D.) 

Z.EAB=:Z.  CAH.    (Each  =  a  rt.  Z  +  Z  CAB.) 

.'.  A  ABE  ^  A  CAH.  (V.  1.) 

A  ABE=:i  the  D  on  b} 

(Having  the  same  base  EA  and  =  altitudes.) 
[The  area  of  a  A  =  ^  the  product  of  b  •  h.]  (XIII.  2  a.) 

A  CAH  =^^  the  nnA-L} 

(Having  the  same  base  AH  and  =  altitudes.)' 

.-.  i  D  on  6  =  ^  the  □  A-L.  (Ax.  1.) 

.-.  the  D  on  6  =  the  cz]  A-L.  (Ax.  2.) 

Similarly,  we  may  show  that  the  D  on  a  =  □  L-B. 

.'.  the  D  on  a  +  the  D  on  6  =  □  A-L  -\-  □  L-B  =  D  on  c. 

Q.E.D. 

XIV.  1  a.  The  square  on  a  leg  of  a  right  triangle 
equals  the  difference  between  the  square  on  the  hypote- 
nuse and  the  square  on  the  other  leg. 


1  Draw  altitude  in  A  ABE  from  BtoEA;iJiA  CAH  from  O  to  AH. 


XIV.     PYTHAGOREAN  GROUP 


151 


XIV.  2.  In  any  triangle,  the  square  on  a  side  opposite 
an  acute  angle  equals  the  sum  of  the  squares  on  the 
other  tico  sides,  diminished  by  twice  the  rectangle  of 
either  of  these  sides  and  the  projection  of  the  other 
upon  it 


f 

K 

c 

Kyp.  If,  in 
the  A  ABC, 
Z.C  is  acute ; 
CT,  the  pro- 
jection of  b  on 
a,  and  CM,  the 
projection  of  a 
on  b. 


H     K 
Ck)nc. :  then  D  onc=D  ona+D  on6  —  2.nofa«  GT,  or 
2'niotb'CM. 

Dem.     Draw  the  altitudes  and  extend  them  to  meet  sides  of 

the  D's  as  shown  in  the  figure.     These  altitudes  must  pass 

through  Jf  and  T.  (Def.  of  projection.) 

If,  as  in  figure  for  XIV.  1,  CH  and  BG  were  drawn,  A  CAH 

would  be  ^  A  BAG.     ...  □  A-K=n2  G-M.  (Ax.  3.) 

Similarly,  C=i  5- JT  =  □  jE7-J5.    . 

Again,  if  BQ  and  AF  were  drawn, 

ABCq^h.AGF. 
.'.  □  M-Q  =  Z3T-F.  (Ax.  3.) 

cz]  A-K+  □  B-K  =  □  G-M  +  □  E-B,  (Ax.  2.) 

=  D  on  6  -  □iW-Q  +  D  on  a  -  □  T-F, 
or  (since  □  T-F  =  □  M-Q) 

=  non6  +  nona-2a  M-Q, 

=  non6  +  nona  —  2aof6  and  CM, 

=  Don6  +  nona  —  2nofa  and  CT. 

Q.E.D. 


152 


THE  ELEMENTS  OF  GEOMETRY 


XIV.  3.  In  any  obtuse  triangle,  the  square  on  the  side 
opposite  the  obtuse  angle  equals  the  sum  of  the  squares 
on  the  other  two  sides,  increased  by  twice  the  rectangle 
of  either  of  these  sides  and  the  projection  of  the  other 
upon  it 


Hyp.    If,  in 
the    A  ABC, 

ZC is  obtuse; 
CT  the  pro- 
jection of  b 
on  a,  and  CM 
the  projection 
of  a  on  b, 


Cone. :  then  nonc  =  nona  +  non6  +  2«aof6  and  CM, 
=  D  on  a  +  D  on  &  +  2  .  □  of  a  and  or. 

Dem.     Draw  the  altitudes  of  A  ABC  and  produce  them  as  in 
the  figure ;  also  draw  AF  and  BG. 

If  EB  and  CH  be  drawn,  □  E-M  may  be  proved  =  □  A-K. 
[Each  □  being  twice  one  of  the  ^  A  ABEsiRd  ACH (why  ^  ?).] 

Similarly,        -  □  B-S  =  □  B-IC 

(Draw  CL  and  AQ  and  give  remainder  of  proof.) 
.-.  cDA-K-h\Z}B-K=c3E-M-\-C3B-S. 
That  is,  the  D  on  c  =  □  E-M  +  □  B-S. 

But  nDE-M=  the  D  on  6  +  □  O-M, 

and  □  B-S  =  the  D  on  a  +  □  C-S. 

.-.  the  D  on  c  =  the  D  on  6  +  the  D  on  a  +  □  G-M+  □  C-S. 
But       □  G-M  =  □  C-S,  -.'  A  ACF  ^  A  GGB.         (Why  ?) 
But    ^G-S^CT'CFov  CT'a,B.xidin^G-M:=b'CM. 
. .  G  on  c  =  D  on  6  4-  D  on  a -f  2  •  □  of  a  and  CT, 

=  Don6  +  Dona-|-2-aof6  and  CM. 

Q.E.D. 


XIV.    PYTHAGOREAN  GROUP  153 

Ex.  2.  The  radius  of  a  circle  is  r.  A  tangent  of  length  1  is  drawn  to 
this  circle. 

Draw  the  hypotenuse  and  find  an  expression  for  its  length.  What  is 
the  locus  of  the  extremity  of  this  tangent  ? 

Ex.  3.  A  ladder  50  ft.  long  just  reaches  the  top  of  a  wall  40  ft.  high. 
If  the  ground  be  level,  how  far  is  the  foot  of  the  ladder  from  the  wall  ? 

Ex.  4.  (a)  The  length  of  a  chord  is  12  ft.  What  is  its  distance  from 
the  center  of  a  circle  whose  radius  is  20  ft.  ? 

(6)  If  the  length  of  a  chord  be  2  a,  how  far  is  this  chord  from  the 
center  of  a  circle  of  radius  r  ? 

Ex.  5.  What  is  the  length  of  a  chord  in  a  circle  of  radius  r,  and  at  a 
distance  d  from  the  center  ? 

Ex.  6.  Find  the  area  of  the  cross  section  of  a  ditch,  the  section  being  an 
isosceles  trapezoid  20  ft.  wide  at  the  bottom,  30  ft.  wide  at  the  top,  and 
each  slope  25  ft.  from  top  to  bottom. 

Ex.  7.  If  a  public  square  is  200  yds.  on  a  side,  how  much  is  gained  by 
crossing  the  square  from  corner  to  corner  on  a  diagonal  walk  instead  of 
using  the  sidewalk  around  the  square  ? 

Ex.  8.   Find  the  side  and  area  of  the  □  inscribed  in  a  O  of  radius  a. 

Ex.  9.  The  sides  of  a  triangle  are  6,  8,  and  10,  respectively.  What  kind 
of  triangle  is  it  ?     Why  ? 

Ex.  10.  The  sides  of  a  A  are  5,  7,  and  10,  respectively.  What  kind  of 
A  is  it  ? 

Ex.  11.    If  the  three  sides  of  a  triangle  are  given,  we  may  find : 

(1)  The  lengths  of  the  projections  of  two  of  the  sides  on  the  third  side; 

(2)  The  length  of  the  altitude  to  the  third  side ; 

(3)  The  area  of  the  triangle, 

in  the  following  manner  : 

(1)  Let  the  sides  of  the  triangle  be  8,  11,  and  14. 

Let  X  and  y  be  the  projections  on  14  of  11  and  8,  respectively. 

x  +  y  =  14. 
x^  —  y2  =  57        (XIV.  1  and  •.•  the  altitude  is  common  to  both  A.) 
14(ic  -y)z=67,orx-y  =  f|. 

Knowing  x  ■{■  y  and  x  —  y,  we  may  find  x  and  y. 

(2)  Knowing  either  x  or  y,  we  may  find  the  altitude  by  XIV.  1  a. 

(3)  Knowing  the  base  and  altitude,  we  have  the  area. 


154  THE  ELEMENTS   OF   GEOMETRY 


XIV.     SUMMARY  OF  PROPOSITIONS  IN  PYTHAGOREAN 

GROUP 

1.  If  a  triangle  is  right,  the  square  on  the  hypotenuse 
equals  the  sum  of  the  squares  on  the  other  two  sides. 

a  The  square  on  the  side  of  a  right  triangle  equals 
the  difference  between  the  square  on  the  hy- 
potenuse and  the  square  on  the  other  side. 

2.  In  any  triangle,  the  square  on  a  side  opposite  an 
acute  angle  equals  the  sum  of  the  squares  on  the  other 
tivo  sides,  diminished  by  tivice  the  rectangle  of  either  of 
these  sides  and  the  projection  of  the  other  upon  it. 

3.  In  any  obtuse  triangle,  the  square  on  the  side  oppo- 
site the  obtuse  angle  equals  the  sum  of  the  squares  on  the 
other  two  sides,  increased  by  twice  the  rectangle  of  either 
of  these  sides  and  the  projection  of  the  other  upon  it. 


XIV.     PYTHAGOREAN   GROUP  155 

Ex.  12.  Having  the  length  of  two  sides  of  a  A  and  of  the  altitude  to 
the  third  side,  how  do  you  find  the  length  of  the  third  side  ? 

Ex.  13.  If  each  side  of  an  equilateral  triangle  equals  10,  show  that  the 
altitude  is  V75  =by/S,  and  the  area  25  V3. 

Ex.  14.    In  an  equilateral  triangle,  if  one  of  the  sides  is  a  and  the  altitude 

is  h,  show  that  h=-  V3. 
2 

Ex.  15.   What  is  the  area  of  an  equilateral  triangle  whose  side  is  20  ft.  ? 

Ex.  16.   Prove  that  in  any  triangle,  ABC,  if  p  and  q  are  the  segments 
of  c  made  by  an  altitude  on  c,  and  a  and  b  are  the  remaining  sides,  then 
a  +  b  : p  +  q  : : p  —  q  :  a  —  b. 

Ex.  17.  Prove  that  if  a  line-segment  makes  an  Z  of  60°  with  the  base  of 
projection,  the  length  of  the  projection  is  one  half  the  original  line-segment. 

Ex.  18.  If  a  line-segment  of  length  a  makes  an  angle  of  30°  with  the 
base  of  projection,  how  long  is  the  projection  of  a  ? 

Ex.  19.  The  sides  of  a  trapezoid  are  12,  32,  12,  and  40,  respectively. 
Find  its  area. 

Ex.  20.  An  extension  ladder,  75  ft.  long,  just  reaches  a  window  60  ft. 
from  the  ground.  How  far  is  the  foot  of  the  ladder  from  the  side  of  the 
building  ? 

Ex.  21.  The  sides  of  a  right  triangle  are  three  consecutive  integral 
numbers.     What  is  the  length  of  the  hypotenuse  ? 

Ex.  22.  What  is  the  length  of  the  diagonal  of  a  rectangular  floor  24  ft. 
by  22  ft.  ? 

Ex.  23.  The  diagonal  of  a  rectangle  is  2.9  ft.  ;  the  perimeter  is  8.2  ft. 
What  is  the  length  of  the  rectangle  ? 

Ex.  24.  The  hypotenuse  of  a  right  triangle  is  58  ft. ;  one  leg  is  42  ft. 
What  is  the  length  of  the  altitude  to  the  hypotenuse  ? 

Ex.  25.    The  sides  of  a  triangle  are  5,  6,  and  7. 

Find  the  segments  of  each  side  made  by  the  altitude  upon  it. 

Ex.  26.  Draw  the  projection  of  the  line-segment  on  the  base  of  pro- 
jection in  each  of  the  following  cases : 

1.  The  line-segment  above  the  base  of  projection. 

2.  The  line-segment  meeting  the  base  of  projection. 

3.  The  line-segment  intersecting  the  base  of  projection. 

Ex.  27.  If,  in  a  scalene  triangle,  a  median  is  drawn,  prove  that  one  of 
the  angles  it  forms  with  the  side  is  acute  and  the  other  is  obtuse. 


156  THE   ELEMENTS  OF  GEOMETRY 

Ex.  28.  Hence,  prove  that  in  any  triangle  the  sum  of  the  squares  on  two 
sides  equals  twice  the  square  on  half  the  third  side  plus  twice  the  square 
on  the  median  to  that  side.     (XIV.  2  and  3,  and  combine  by  addition.) 

Ex.  29.  Prove  that  in  any  triangle  the  difference  of  the  squares  on  any 
two  sides  equals  twice  the  rectangle  of  the  third  side  and  the  projection  of 
the  median  on  that  side.     (Use  XIV.  2  and  3 ;  combine  by  subtraction.) 

Ex.  30.  Verify  by  means  of  Ex.  28  the  theorem  that  the  square  on  the 
hypotenuse  equals  the  sum  of  the  squares  on  the  two  sides. 

Ex.  31.   Prove  that  the  sum  of  the  squares  on 
the  four  sides  of  a  parallelogram  equals  the  sum  of        ---'''^''/2^^^^^-''<C/ 
the  squares  On  its  diagonals.  '^" 

Ex.  32.  Prove  that  in  a  right  triangle  twice  the  sum  of  the  squares  of 
■the  medians  equals  three  times  the  square  of  the  hypotenuse. 

Ex.  33.   If,  in  rt.  A  ABC  and  ABC,  M  is  the  mid-      ^ 
point  of  hypotenuse  AB,  what  does  AC'^  +  BC^  equal  ? 
What  does  AC'^  +  BC'^  equal  ? 


M  B 


Ex.  34.   Deduce  from  Ex.  28  that  CM^  (or  Cld^)  =  AM\ 


Ex.  35.   If  the  point  C  moves  so  that  it  is  always  the  vertex  of  a  right 
gle  whose  sides  pass  th 
what  constant  quantity  ? 


angle  whose  sides  pass  through  A  and  B,  then  CA^  +  CB^  always  equals 


Ex.  36.  What  is  the  center  and  what  the  radius  of  the  circle  that  is  the 
locus  of  all  points,  the  sum  of  the  squares  of  the  distances  from  which 
points  to  A  and  ^  is  a  square  on  the  join  of  A  and  B  ? 

Ex.  37.    In  an  oblique  A  ABC,  M  is  th-e  mid-point  of  AB. 

What  does  a'^  +  b^  equal  ?     (v.  Ex.  28.) 

What  does  BC"2  +  ^C"2  equal  ?  ^' 

Ex.  38.   If  point  C  in  the  annexed  diagram  moves  so 
that  a^  4-  6^  always  equals  a  given  square,  say  Q%  show 

that  CM^  =  ^-AM^. 

2  _  _  _ 

Ex.  39.    Construct  a  square  =  3L  _  AM^. 

Ex.  40.  What  is  the  center  and  what  the  radius  of  the  circle  that  is  the 
locus  of  all  points,  the  sum  of  the  squares  of  the  distances  from  which 
points  to  A  and  B  equals  a  given  square  Q^  ? 


^^. 


XIV.     PYTHAGOREAN   GROUP  157 

Ex.  41.   If,  in  the  annexed  diagram,  G  is  the  inter-         L 
section  of  the  medians  of  the  A  ABC^  in  what  ratio  does        /}  \^ 
G^  divide  Oilf? 


/    X 


Ex.  42.    If  L  is  any  point  in  the  plane,  prove  that  /     j       M      \ 

Proof.    Let  J2"  be  mid-point  of  OG^.  '/ 

U^  -f  LB^  =  2  AM^  +  2  LM^.     (Why  ?) 
LC^  +  ^^  =  2  LH^  +  2  GM\     (Why  ?) 


(Add,  and  combine  terms.) 

Ex.  43.  If  L  moves  so  that  the  sum  of  the  squares  of  its  distances  from 
A^  B,  and  C  =  a  given  square  ;  that  is,  so  that  LA^  -f-  LB^  +  LC^  equals, 
say,  Q^,  what  is  the  center  and  what  the  radius  of  the  locus  circle  ? 

Ex.  44.  The  sum  of  the  squares  of  the  medians  of  a  triangle  equals  three 
fourths  of  the  sum  of  the  squares  of  the  three  sides.     Prove. 

If  the  area  of  a  triangle  is  112  sq.  ft.  and  its  altitude  is  4  ft.  : 
Ex.  45.    What  is  the  length  of  its  base  ? 

Ex.  46.  What  is  the  length  of  the  median  to  the  base  if  its  projection 
on  the  base  is  3  ft.  ? 

Ex.  47.   What  does  the  sum  of  the  squares  on  the  two  sides  equal  ? 
Ex.  48.    What  does  the  difference  of  the  squares  on  the  two  sides  equal  ? 
Ex.  49.    What,  then,  are  the  lengths  of  the  two  sides  ? 

Ex.  50.  Draw  a  trapezium,  its  diagonals,  and  the  mid-join  of  the  diago- 
nals. Prove  that  the  sum  of  the  squares  on  the  four  sides  equals  the  sum 
of  the  squares  on  the  diagonals  plus  four  times  the  square  on  the  mid-join. 
(Euler's  Theorem.)     (Use  Ex.  28.) 

In  the  accompanying  figure,  A  ABL  is  right-angled      q j?    E 

at  L.     A-K  is  a  square  on  AB,  L-F  on  AL^  L-E  on  ~^^Z^^  V 

BL.     Prove  the  following  relations :  4  ^""^ ^  ■*-  M 

Ex.51.    (\)  LAFa^l\ABL.  \  \ 

Ex.52.    (2)  AOBQ^AJMK.  \  e^^ 

Ex.  53.    (3)  4-side  JKBL  =  A BEK.  \  ^,^^\\ 

Ex.  54.    (4)  Hence,  by  means  of  this  figure,  prove     F    6  Q  M 

XIV.  1. 

Ex.  55.  (5)  Also  prove  XIV.  1  by  showing  that  the  remainder  obtained 
by  subtracting  from  the  large  square,  the  sum  of  the  medium  and  small 
squares,  is  equal  to  that  obtained  by  subtracting  from  the  large  square,  the 
square  on  the  hypotenuse  AB. 


158 


THE  ELEMENTS  OF  GEOMETRY 


Ex.  56.  If,  in  the  figure  for  Ex.  40,  A  and  B  are  the  centers  of  circles  of 
radii  r  and  r',  respectively,  what  is  the  locus  of  a  point  that  moves  so  that 
the  sum  of  the  squares  of  the  tangents  from  the  moving  point  to  the  two 
given  circles  equals  a  given  square  ?    (Use  Ex.  40  and  XIV.  1  a.) 

Ex.  57.  Prove  that  the  external  common  tangent 
to  two  tangent  circles  is  a  mean  proportional  to  the 
diameters  of  the  two  circles. 

(Draw  line  as  in  the  annexed  diagram.  Hypote- 
nuse of  rt.  A  =  r  +  r' ;  short  leg  =  r  —  r'.  Use 
XIV.  1  a.) 

Ex.  58.  Find  the  locus  of  the  extremities  of  lines  that  all  pass  through 
the  same  point  ilf  on  a  given  line  AB,  and  have  the  same  projection  on  AB. 


Ex.  59.  If  squares  are  drawn  on  the  three 
sides  of  rt.  A  ABC,  prove  that  BA  JL  CL.  £; 

Ex.  60.  Prove  that  if  KF±  EA  in  this  figure, 
I^AKF^ISABC. 

Ex.  61.  In  the  same  figure  prove  that  the 
area  of  A  ir^i^=  the  area  of  A  BBL. 


Ex.  62.  If  the  diagonals  of  a  4-side  are  perpendicular  to  each  other, 
show  that  the  sura  of  the  squares  of  one  set  of  opposite  sides  equals  the 
sum  of  the  squares  of  the  other  set. 

Ex.  63.  The  4-side  A-J  is  any  parallelo- 
gram on  AC,  and  4-side  C-L  is  any  paral- 
lelogram on  CB.  If  LH  and  BJ  are  produced 
to  intersect  at  F,  and  then  if  a  O  A-M  is  con- 
structed on  AB  as  a  base,  whose  sides  are  equal 
and  parallel  to  FC,  prove  : 

(1)  By  Ax.  7,  that  join  i^'itf  must  fall  on  FL. 

(2)  By  XIII.  1  c,  Sch.  3,  th&t  O  L-C  =  O  B- F. 

(3)  Th&t  a  B-K=  a  B-F. 

(4)  Why,  then,  is  O B-K  =  O L-C  ? 
Show  in  a  similar  manner  that  O  A~K=  O  A-J. 
Prove  therefore  that  O  A-M  =  O  L-C  +  O  B-C . 


Ex.  64. 
Ex.  65. 
Ex.  66. 


By  means  of  the  foregoing,  prove  that  M/LACB  is  a  right  angle, 


then  the  square  on  AB  =  the  square  on  ^C  +  the  square  on  BC. 


XIV.     PYTHAGOREAN   GROUP 


159 


PROBLEMS 

Prob.  I.     To  construct  a  square  equal  to  the  sum  of 
two  given  squares. 
Given,     a  and  6,  the  sides  of  2  D's.  h 

Required.     To  construct  a  D  equal 
to  a"  +  h\ 

[Construction  is  left  to  the  student.] 

Prob.  II.     To  construct  a  square  equal  to   twice   a 
given  square. 

[Construction  is  left  to  the  student] 

Prob.  III.    To  C07istruct  a  square  equal  to  the  sum  of 
three  given  squares. 

Given,     a,  b,  and  c,  the  sides 
of  3  D's. 


Required.     To  construct  a  D 
equal  to  d^  -\- b^  -\-  cl 

[Construction  is  left  to  the  student.! 

Prob.  IV.     To  construct  a  square  equal  to  the  differ- 
ence  of  two  given  squares. 


Given.     The 

sides  a  and  b 
of  two  squares. 


Required.     To  construct  a  square  equal  to  their  difference. 
Const.     Const,  is  left  to  the  student. 


XV.     GROUP   ON   SIMILAR   FIGURES 

DEFINITIONS 

Two  Polygons  are  Similar  when  the  angles  of  the  first  are 
respectively  equal  to  the  angles  of  the  second,  and  the  homolo- 
gous sides  are  proportional. 

Homologous  Sides  of  Similar  Polygons  are  the  sides  connecting 
the  vertices  of  corresponding  angles. 

Homologous  Lines  are  lines  similarly  drawn  in  the  two 
polygons. 

Illustrations  :  Homologous  lines  in  similar  triangles  are 
the  corresponding  medians,  altitudes,  angle  bisectors,  etc. 

Homologous  lines  in  similar  polygons  are  the  corresponding 
diagonals,  radii  of  ciroumscribe'i  and  of  inscribed  circles,  etc. 

I'he  Ratio  of  Similitude  of  any  two  similar  figures  is  the  ratio 
of  any  two  homologous  lines  of  the  figures. 

Two  or  more  figures  are  in  Perspective,  when  they  are  so 
placed  that  the  joins  of  Corresponding  Points  concur. 

In  polygons  the  corresponding  points  are  usually  vertices. 

A  Center  of  Similitude  is  a  point  of  concurrence  of  corre- 
sponding joins  of  similar  figures  in  perspective. 

Draw  4 

Ex.  1.   Two  similar  triangles. 
Ex.  2.   Two  triangles  that  are  not  similar. 

Ex.  3.  Two  quadrilaterals  that  are  mutually  equiangular  but  not  similar. 
Ex.  4.    Two  quadrilaterals  that  have  the  sides  of  the  one  proportional 
to  the  sides  of  the  other,  but  are  not  similar. 
Two  triangles  are  similar.     Show  that 

Ex.  5.    If  the  first  be  isosceles,  the  second  will  also  be  isosceles. 
Ex.  6.   If  the  first  be  equilateral,  the  second  will  also  be  equilateral. 
Ex.  7.    If  the  first  be  right,  the  second  will  also  be  right. 

160 


M     E                       ^F 

ir 

/''^^^--<'''       \ 

A*"                                                 "      " 

D 

XV.     SIMILAR  FIGURES  161 

XV.  1.  If  a  line  is  parallel  to  one  side  of  a  triangle, 
it  divides  the  other  two  sides  proportionally. 

Hyp.  If, 
in  A  ABC, 
MK  II  AB, 

and  cuts  CA 
in  E  and  CB 

Cone. :  then  CB.CF::  CA  :  CE. 

Dem.     Draw  AF  and  BE. 

The  A  EBF  and  EEC  have  the  sam-e  altitude ;  for  a  per- 
pendicular dropped  from  E  to  the  line  CFB  would  be  the 
altitude  of  each.  (Def.  of  altitude  of  A.) 

Similarly,  the  A  AFE  and  EFC  have  the  same  altitude. 

.-.  AEFC.AEBF: :  CF  :  FB.      (1) 

[A  with  equal  altitudes  are  to  each  other, etc.]  (XIII.  1  c,  Sch.  2.) 
A  EFC :  A  AFE  : :  CE  :  EA.     (2)     (Same  reason.) 

The  A  EBF  and  AFE  have  the  same  base  EF. 

Their  altitudes  are  equal ;  for  perpendiculars  dropped  from 

A  and  B  on  MK  would  be  opposite  sides  of  a  rectangle,  and 

equal.  (VI.  1  a.) 

.♦.  A  EBF  =  A  AFE.       (XIII.  1  c,  Sch.  3.) 

.-.  from  proportions  (1)  and  (2), 

CF:FB=CE:EA.  (Ax.  1.) 

Taking  this  proportion  by  composition, 

CF  +  FB:  CF:  iCE  +  EA:  CE,  (XI.  1  a.) 

or  CB:CF::CA.CE. 

Q.E.D. 

Ex.  8.  The  sides  of  a  triangle  are  3,  5,  7.  ■  Find  the  sides  of  three  ^  each 
of  which  shall  be  similar  to  the  first  and  shall  have  one  side  equal  to  10,5. 


162  THE  ELEMENTS  OF  GEOMETRY 

XV.  1  a.  Conversely.  If  a  line  divides  two  sides  of  a 
triangle  proportionally,  this  line  is  parallel  to  the  third 
side. 


Hyp.     If,  in  the 
A  ABC,      EF      is 
drawn  so  that 

CB'.CFi.CA.CE, 

y             ^V^ 

aL 

\.. 

Cone. :  then 

EFWAB.  * 

Dem.     If  EF  if  AB,  draw  EL  1!  AB. 

Then 

CB 

CL::CA:  CE, 

(XV.  1.) 

and 

CB 

CF::CA:CE. 

(Hyp.) 

.*. 

CB 

:CF::CB:CL. 

(Ax.  1.) 

,., 

CF=CL. 

.-.  F  must  be  the  same  point  as  Z;  i.e.  EF  must  coincide 

with  EL.  .     (Ax.  6.) 

.-.  ^F  must  be  II  AB, 

Q.E.D. 

ScH.  Proportional  division  of  the  sides  of  a  triangle  may 
take  place  in  three  different  ways. 

Thus,  in  the  proof  of  XV.  1,  we  have  established  two  pro- 
.portions  involving  CB  and  CA  and  their  segments,  viz. : 

CF:FB::CE:EA,  (1) 

CB.CF.iCA:  CE.  (2) 

We  may  also  obtain  from  XV.  1  by  composition  (XI.  8), 

CF  +  FBxFB-.'.CE  +  EA:  EA, 

or  CB'.FB'.'.CA'.EA.  (3) 

Ex.  9.  The  sides  of  a  right  triangle  are  3,  4,  5.  The  hypotenuse  of  a 
similar  right  triangle  is  60.    Find  the  other  sides  of  the  second  right  triangle. 

Ex.  10.  The  sides  of  a  triangle  are  7,  10,  12,  and  the  longest  side  of  a 
similar  triangle  is  18.     Find  the  remaining  sides  of  the  second  triangle. 


XV.     SIMILAR   FIGURES  163 

XV.  2.  If  in  two  triangles  the  angles  of  the  one  are 
equal  to  the  angles  of  the  other,  the  sides  of  the  triangles 
a7'e  proportional,  and  the  triangles  are  similar. 

Hyp.     If,  in  the 
A  ABC  and  EFG, 

AA  =  ZE, 

and  ZC=/.G, 

Cone. :  then  CB.GF.iCA  :  GE  :  :  AB  :  EF, 

and  AABC'^AEFG 

Dero.  We  may  place  the  A  EFG  on  the  A  ABO  so  that  it 
takes  the  position  CE'F'.     (Why  ?) 

ZCE'F'  =  ZA.  (Hyp.) 

.-.  E'F'  II  AB.  (Def.  of  lis.) 

Similarly,  by  making  E  coincide  with  A,  we  may  show  that 

CA^AB 
GE     EF' 

**  GF      GE     EF'  ^     '    •) 

,\  A  ABC  ^  A  EFG.  (Def.  of  ~  A.) 

Q.E.D. 

XV.  2  a.  Two  right  triangles  are  similar  if  an  acute 
ajigle  of  one  equals  an  acute  angle  of  the  other. 

Two  4-sides  are  similar.     Show  that 

Ex.  11.   If  the  first  be  a  trapezoid,  the  second  will  also  be  a  trapezoid 
Ex.  12.    If  the  first  be  a  parallelogram,  the  second  will  also  be  a  par- 
allelogram. 


164         THE  ELEMENTS  OF  GEOMETRY 

XV.  3.  If  in  two  triangles  the  sides  of  the  one  are 
respectively  proportional  to  the  sides  of  the  other,  the 
angles  of  the  first  are  equal  to  the  angles  of  the  second, 
and  the  triangles  are  similar, 

Q  0 

Ryp.     If,  in  the 
AABCsiVidEFGy 

CB^CA^AB 
OF      GE     EF' 

Cone:    then    Z.A=iZ.E]    AB 

AABC'^AEFG. 

Dem.   Lay  off  OF'  =  GF,  and  CE'  =  GE;  draw  E'F'. 
Substituting  CE'  and  CF'  for  their  equals  GE  and  GF  in 
the  given  proportion,  we  have 

CB^CA 
CF'     CE'' 

.-.  E'F'  II  AB.  (XY.  1  a.) 

.-.  Z CE'F' ^Z.A',  A CF'E'  =  Z  J5.  (Def .  of  lis.) 

.-.  A  CE'F'  ^  A  ABC.  (XV.  2.) 
.    CA       AB         ,^_^    ^ 


t.e. 


But 


But 


"  CE'     E'F'' 

[^UGl.  ( 

01  -^  pi 

oiygons.; 

CA       AB 
GE     E'F' 

(CE' 

=  GE 

;  Const.) 

CA     AB 
GE     EF 

(Hyp.) 

.   AB      AB 
"  EF     E'F'' 

(Ax.  1.) 

.'.  E'F'=:EF. 

ACE'F'^AEFG, 

(Y.3.) 

A  CE'F' r^  A  ABC. 

(V. 

line  9  of  Dem.) 

,  AEFG^AABC. 

Q.E.D. 


XV.     SIMILAR   FIGURES  165 

XV.  4.  If  two  sides  of  one  triangle  are  proportional 
to  two  sides  of  a  second,  and  the  included  angles  are 
equal,  the  triangles  are  similar. 

G 
Hyp.     If,  in  the 

AABC2A\^EFG, 

CA:GE::CB:GF, 

Cone:  then  AABC'^AEFG. 

Dem.     Place  A  EFG  on  A  ABC  so  that  it  will  take  the  posi- 
tion E'F'C.  ^^  :CF'::CA:  CE'.  (Hyp.) 
.-.  E'F'  II  AB.                            (XV.  1  a.) 
/.  Z  CE'F'  =  Z  ^  and  Z  CF'E'  =  Z.B.     (Def .  of  lis.) 
Z.C=Z.G. 

.-.  A  ABC  -  A  EFG.  .    (XV.  2.) 

Q.E.D. 

ScH.  If  two  right  triangles  have  the  legs  of  the  first  pro- 
portional to  the  legs  of  the  second,  the  two  right  triangles  are 
similar. 

Two  4-sides  are  similar.     Show  that 

Ex.  13.   If  the  first  be  a  rectangle,  the  second  will  also  be  a  rectangle. 

Ex.  14.  If  the  diagonals  of  the  first  be  equal,  the  diagonals  of  the 
second  will  also  be  equal. 

Ex.  15.    If  the  first  be  cyclic,  the  second  will  also  be  cyclic. 

Ex.  16.  If  the  first  be  circurascriptible,  the  second  will  also  be  circum- 
scriptible. 

Ex.  17.  If  the  diagonals  of  the  first  be  at  right  angles,  the  diagonals  of 
the  second  will  be  at  right  angles. 

Ex.  18.  If  two  isosceles  triangles  have  their  vertex  angles  equal,  the 
triangles  are  similar. 

Ex.  19.  If  in  two  isosceles  triangles  a  leg  and  the  base  of  one  be  propor- 
tional to  a  leg  and  the  base  of  the  other,  the  triangles  are  similar. 


1(J0  THK  ELEMENTS  OF  GEOMETRY 

XV.  6.  If  two  polygons  are  similar,  they  may  he 
divided  into  the  same  number  of  triangles,  similar  in 
corresponding  pairs. 

Hyp.     If  the 

two  polygons 
A-F  and  O-K 
are  similar, 

Cone. :  then  A-F  and  O-K  may  be  divided  into  the  same 
number  of  triangles,  similar  in  corresponding  pairs. 

Dam.     Draw  the  homologous  diagonals  ACj  AE,  GI,  OJ. 
The  polygons  are  thus  divided  into  the  same  number  of 
triangles. 

To  prove  these  triangles  similar  in  corresponding  pairs. 

A  ABC  -  A  GHL  (XV.  4.) 

AACE'^AGIJ, 

For                     Z  BCE  =  Z  HIJ.  (Def .  -  polygons.) 

Z.  BCA  =  Z  HIG.  (Homologous  Z  of  -^  A.) 

.-.  ZACE  =  Z  GIJ.  (By  subtraction;  Ax.  2.) 


•) 


Again,  ^=/^^=^.  (Hom.sidesof-^A 

GI      \I£I J      Ij 

.'.  A  ACE  ~  A  GIJ.  (XV.  4.) 

Similarly  for  the  remaining  pairs  of  triangles. 

.-.  the  triangles  are  similar  in  corresponding  pairs. 

Q.E.D. 

Ex.  20.  Show,  from  Ex.  19,  by  drawing  the  altitudes  of  the  isosceles 
triangles,  that 

If  in  two  right  /^  the  hypotenuse  and  a  leg  of  the  first  are  proportional 
to  the  hypotenuse  and  a  leg  of  the  second,  the  right  ^  are  similar. 

Ex.  21.   All  equilateral  triangles  are  similar. 


XV.     SIMILAR    FIGURES 


167 


XV.  5  a.  Conversely.  If  two  polygons  maybe  divided 
into  the  same  number  of  triangles,  similar  in  corre- 
sponding pairs  and  similarly  2^lciced,  the  polygons  are 
similar. 

Let  the  student  supply  the  proof,  showing  that : 
(o)  The  homologous  angles  of  the  polygons  are  equal,  using 
addition  where  in  XV.  5  we  have  used  subtraction. 
(&)  The  homologous  sides  are  proportional,  e.g. 

BC:HI::AO:  GI, 
and  CE:IJ:.AC:GI) 

whence,  BO :  HI : :  CE :  JJ,  etc.  (Ax.  1.) 


Ex.  22.  If  two  ZI7  have  an  angle  of  the  first  equal  to  an  angle  of  the 
second  and  the  including  sides  proportional,  the  UJ  are  similar. 

Ex.  23.  From  a  triangle  of  altitude  20  ft.  and  base  60  ft.  a  small  triangle 
is  cut  off  by  a  line  parallel  to  the  base  at  a  distance  of  4  ft.  from  the  vertex. 

What  is  the  area  of  the  trapezoid  remaining  ? 

Find  the  values  that  will  satisfy  the  following  given  conditions : 

Ex.  24.  Find  &,  given  «  =  8,  «  =  3,  s  =  12. 

Ex.  25.  Find  e,  given  6  +  e  =  20,  a  +  s  =  35,  a  =  26. 

Ex.  26.  Find  «,  given  a  =  6  +  e,  &  =  16,  a  +  «  =  25. 

Ex.  27.  Find  &  +  e,  given  6  =  4,  a  =  5,  s  =  1. 

Ex.28.  Find  6,  given/=2&,  c  =  24,  e  =  3.  6,/^<* 

Ex.  29.  Given  6  =  10  ft.,  e  =  4  ft. ,  c  =  35  ft.    Find  /. 

Ex.  30.  Given  &  +  e  =  25  ft.,  c  =  40  ft.,  6  ~  e  =  7  f t. 
Find/. 

Ex.  31.  Given  &  +  e  =  25  ft.,  c  =  35  ft.,  /  =  24.     Find  6  and  e. 

Ex.32.  Given/=2  6,  c  =  30ft.,  p-e  =  3.     Find  &  and  e. 


Ex.  33.  Wishing  to  find  the  breadth  {BX) 
of  a  river,  I  measure  off  the  line  AB.,  and  at  A 
draw  AE  parallel  to  BX,  and  EX,  crossing 
AB  at  C.     Show  how  to  find  the  distance  BX. 


168  THE  ELEMENTS  OF  GEOMETRY 

XV.  6.  If  two   triangles   are    similar,   thefj    may    be, 
placed  in  perspective,  a 

Hyp.    If  _,..- ^' 

are  similar, 

Cone. :  then  they  may  be  so  placed  that  AxA^  BiB,  and  C\  C 
will  concur. 

Dem.     Place  the  triangles  so  that  any  two  pairs  of  sides,  as 
AB  and  AiBi,  BC  and  BiC\,  are  parallel. 

Draw  AAi  and  BBi,  and  let  them  intersect  at  0. 

AABO^AA^BiO.  (XV.  2.) 

.-.  OA  :  OA^::  OB  :  OB^  :  :  AB  :  A^B,. 
Now  A^  :  AiBi  is  the  ratio  of  similitude  of  the  A  ABC  and 
AiBiCi. 

Draw  CCi  and,  if  possible,  let  it  intersect  BBi  in  M. 
A  ABC  =  Z  A^By  Ci  and  Z  DBA  =  Z  OB^A^.     (Def .  of  ^  A.) 
.-.  Z  OBC  =  Z  O^iCi.  (Ax.  2.) 

.\BC\\  B^C^.  (Def.  of  ||,) 

.-.A  MBC  ^  A  MB^  Ci.  (XV.  2.) 

/.  MB  ;  JfBi  ::  BC  :  B^C^.     (Horn,  sides  of '^  A.) 
But  BC :  i5iCi  :  :  ^^  :  ^^^i.  (Hyp.) 

.-.  MB  :  MB^  ::AB:  A^B^  :  :  OB  :  OB^.      (Ax.  1.) 
That  is,  OB  :  OB^  :  :  MB  :  MB^,  (or  by  division) 

OB  -  OBi  {=  BBj)  _MB-  MB^  (=  BB{) 
OB  ~'  MB 

As  BBi  =  BBi,  OB  must  be  =  MB\  that  is,  3f  must  fall  on  O. 
,'.  AiAy  BiB,  and  C-^C  must  concur. 

O.E.D. 

XV.  6  a.  If  two  figures  in  perspective  have  their  sides 
parallel,  they  are  similar. 


XV.     SIMILAR  FIGURES  '  169 

XV.  Gb.  If  two  figures  are  in  perspective,  and  the 
joins  of  homologous  points  are  divided  proportionally 
by  the  figures^  they  are  similar. 

XV.  7.  If  two  sectors  have  equal  central  angles,  th&ir 
arcs  are  proportional  to  the  radii. 

Hyp  if  the 
sectors  AKE 
and  FKL  have 
the    commoii 

JT- X  ^ 

Cone. :  then  arc  ABCE   arc  B'GHL  -. .  KA .  KF. 

Dem.  Divide  AE  into  any  number  of  equal  arcs  AB.  BCy  etc., 
and  draw  the  chords  AB,  BC,  etc.  Join  A,  B,  etc./  to  K,  and 
connect  the  points  of  intersection  with  arc  FL  by  FG,  GH,  etc, 

A  KAB  ^  A  KFG.  (XV.  4.) 

Similarly,  A  KBC  ~  A  KGH,  etc. 

.-.  w-gon  K-ABGE  ~  n-gon  K-FGHL.      (XV.  5  a.) 

AB  +  BC+CE  ^KA  ... 

"  FG-^GH  +  HL     KF  ^^ 

(Horn,  lines  of  ~  figures.) 

If,  now,  the  arcs  AB,  BC,  etc.,  be  bisected,  and  new  polygonal 
sectors  be  formed  having  double  the  number  of  angles,  these 
sectors  will  still  be  similar,  and  proportion  (1)  will  continue  to 
be  true,  no  matter  how  far  the  process  be  carried. 

But  by  continuing  the  process  of  bisection,  we  may  make 
the  polygonal  sector  approach  as  near  as  we  please  to  the  circu- 
lar sectors,  which  are  therefore  the  limits  of  the  similar  polygons. 

.*.  at  the  limits  we  have 

2iYQABCE ^KA  ^j  ^. 

qxgFGHL     KF  ^     '     '^ 

Q.E.D. 


170         THE.  ELEMENTS  OF  GEOMETRY 

XV.  8.  Regular  polygons  of  the  same  number  of  aides 
are  similar. 

Hyp.    If  two 

regular      poly- 
gons have  the       ^^  ^ 
same     number 
of  sides, 


Cone. :  then  the  polygons  are  similar. 

Dem.  Let  M  and  M'  be  the  polygons,  71  being  the  number  of 
sides  in  each. 

Any  angle  of  M equals  any  angle  of  M'  =    ^~    rt.  A. 

n 

[In  a  regular  polygon  each  int.  Z  =    ^~    rt. zS.]     (III.  3  a.) 

.'.  Jlf  and  M'  are  mutually  equiangular. 

Again,  AB  =  BC;  A'B  =  B'Q',  etc.    (Def.  of  reg.  polygons.) 

.:  AB:  A'B'  ::B0:  B'C,  etc. ;  (Ax.  3.) 

i.e.  the  sides  of  M  are  proportional  to  the  sides  of  M'. 

.'.  ifcr~  M'.  (Def.  of  '^  polygons.) 

Q.E.D. 

Ex.  34.  In  ^CLF  and  CAB,  ZCLF=ZB,  and  ZCFL  is  supple- 
mental to  Z  CAB.  CL  and  BA  are  produced  to  intersect  at  B,  and  LF  is 
produced  to  meet  CB  at  Q. 

Prove  that  A  CBB  is  isosceles. 

Ex.  35.    Prove  that  CL  :  CB  : :  CF :  CA. 

Ex.  36.    Prove  that  LF:  FQ-.-.BA:  AB. 

Ex.  37.    Prove,  then,  that  LQ  :  BB  :  :  CF :  CA. 

Ex.  38.    If  LF.BA..FQ:  AB  and  if  BL  and 
AF  meet  at  C  while  LQ\\  BB,  prove  that 
GF'.CA'.-.LFxBA. 

Ex.  39.   Similarly,  if  QB  and  AF  meet  at  E,  show 
that^i^:£'^::  FQ.AB. 

Ex.  40.    Show,  then,  that  CF.CA-.iEF:  EA. 

Ex.  41.  Take  the  last  proportion  by  inversion  and  then  by  division  and 
show  that  EF  =  CF. 


XV.     SIMILAR   FIGURES  171 

XV.     SUMMARY  OF  PROPOSITIONS  IN   THE  GROUP 
ON   SIMILAR  FIGURES 

1.  If  a  line  is  parallel  to  one  side  of  a  triangle,  it 
divides  the  other  two  sides  proportionally. 

a  Conversely.  If  a  line  divides  tivo  sides  of  a 
triangle  proportionally,  this  line  is  parallel 
to  the  third  side, 

ScH.  Proportional  division  of  the  sides  of  a  triangle  may 
take  place  in  three  different  ways. 

2.  If  in  tivo  triangles  the  angles  of  the  one  are  equal 
to  the  angles  of  the  other,  the  sides  of  the  triangles  are 
proportional,  and  the  triangles  are  similar. 

a  Tivo  right  triangles  are  similar  if  an  acute  angle 
of  one  equals  an  acute  angle  of  the  other. 

3.  If  in  two  triangles  the  sides  of  the  one  are  respec- 
tively proportional  to  the  sides  of  the  other,  the  angles 
of  the  first  are  equal  to  the  angles  of  the  second,  and 
the  triangles  are  similar. 

4.  If  two  sides  of  one  triangle  are  proportional  to  tivo 
sides  of  a  second,  and  the  included  angles  are  equal,  the 
triangles  are  similar. 

ScH.  If  two  right  triangles  have  the  legs  of  the  first  pro- 
portional to  the  legs  of  the  second,  the  two  right  triangles  are 
similar. 

5.  If  two  polygons  are  similar,  they  may  he  divided 
into  the  same  number  of  triangles,  similar  in  corre- 
sponding pairs. 


172 


THE  ELEMENTS  OF  GEOMETRY 


a  Conversely.  If  two  polygons  may  be  divided 
into  the  same  number  of  triangles,  similar 
in  corresponding  pairs,  the  given  polygons 
are  similar, 

6.  If  tioo  triangles  are  similar,  they  may  be  placed  in 
perspective, 

a  If  two  figures  are  in  perspective,  and  have  their 
sides  parallel,  they  are  similar. 

b  If  two  figures  are  in  perspective,  and  the  joins 
of  homologous  points  are  divided  propor- 
tionally by  the  figures,  they  are  similar. 

7.  If  two  sectors  have  equal  central  angles,  their  arcs 
are  proportional  to  the  radii. 

8.  Regular  polygons  of  the  same  number  of  sides  are 
similar, 

PROBLEM 

Prob.  I.    To  find  a  fourth  proportional    to   three 
given  lines,  ^ 

Given.     The  lines 
a,  b,  and  c,  c //G 

Required.    To  find     ^ 
a     fourth     propor- 
tional to  a,  bf  and  c. 

Anal.     Suppose  x  to  be  the  required  fourth  proportional. 

Then  a:b::c:x.  (See  XV.  1.) 

Const.     On  the  indefinite  line  OL,  take  OA  =  a  and  OC  =  c. 

On  any  intersecting  line  OM,  take  OB  =  b. 

Draw  CX  parallel  to  the  join  AB. 

Then  OX  is  the  required  fourth  proportional,  (XV.  1.) 

O.E  F. 
Proof.     To  be  supplied  by  the  pupil. 


XV.     SIMILAR  FIGURES  178 

Ex.  42.  Prove,  then,  that  if  three  or  more  transversals  intercept  pro- 
portional segments  on  two  parallels,  the  transversals  are  concurrent. 

Ex.  43.  Prove  that  in  a  triangle  the  median  to  the  base  bisects  all  lines 
parallel  to  the  base. 

Ex.  44.  Prove,  then,  that  the  non-parallel  sides  of  a  trapezoid  meet  the 
extended  median  of  the  trapezoid  in  the  same  point. 

Ex.  45.   The  two  points  E  and  F  divide  the  two    q 
sides  oil\ABC  proportionally.    Prove  the  following 
relations : 

(1)  AABE  =  AABF. 

(2)  If,  through  the  point  of  intersection  0,  L3I 
is  drawn  parallel  to  AB,  prove  OL  =  OM. 

(3)  Prove  that  if  GO  is  produced  to  meet  AB  in 
J,AJ=BJ. 

Ex.  46.  In  the  figure  for  the  preceding  theorem, 
state  of  what  points  the  median  CJ  is  the  locus. 
(Two  loci.) 

Ex.  47.  The  sides  of  A  MKL  are  equally- 
inclined  to  those  of  A  ABC;  that  is,  each  when 
produced  makes  an  angle  alpha  with  the  respec- 
tive sides  of  A  ABC.  Prove  the  following  rela- 
tions : 

(1)  That  the  4-side  CFME  is  cyclic. 

(2)  That  the  two  triangles  are  similar. 
Ex.  48.   If  two  angles  of  one  triangle  equal  two  angles  of  another  tri- 
angle, and  the  third  pair  of  angles  are  supplemental,  what  kind  of  triangles 
are  they  ? 

Ex.  49.  Why  can  we  not  have  two  angles  of  one  triangle  supplemental 
to  two  angles  of  another  triangle  ? 

Ex.  50.  Prove,  then,  that  two  triangles  are  similar,  if  the  sides  of  one 
triangle  are  respectively  perpendicular  to  the  sides  of  the  other. 

Ex.  51.  Prove  that  if  the  sides  of  one  triangle  are  parallel  to  the  sides 
of  a  second,  the  triangles  are  similar. 

Ex.  52.  From  any  point  0,  lines  are  drawn  to  the  vertices  of  any  figure 
and  these  lines  are  divided  internally  in  the  ratio  of  two  given  lines.  Prove 
that  the  figure  formed  by  the  joins  of  these  points  of  division  is  similar  to 
the  original  figure. 

Ex.  53.  If  the  lines  in  the  preceding  theorem  are  extended  beyond  the 
point  0  and  divided  externally  in  the  ratio  of  the  two  given  lines,  prove 
that  the  figure  formed  by  the  joins  of  these  points  is  also  similar  to  the 
original  figure. 


174 


THE  ELEMENTS  OF  GEOMETRY 


Ex.  64.  To  divide  a  given  line  into  three  segments,  x,  y,  and  «,  so  that 
x:y::a:b 
tnd  yiz'.iciei 

a,  6,  c,  and  e  being  any  four  given  lines. 

Ex.  65.  If,  in  the  A  ABC,  a  line  CH  is  drawn  to 
any  point  H  in  AB^  and  from  any  point  0  in  CHf 
OA  and  OB  are  drawn,  then 

AOB  :  ABC  I  :  OH :  CK 
Prove. 

Ex.  66.   If,  in  the  above  A  ABC,  lines  AF,  BOj 

CH  are  drawn  through  a  point  0  to  the  sides  a,  6, 

and  c,  respectively,  then 

np      n/2      nrr 


OF      OG      OH 
AF     BG      CH 


Ex.  67.  Prove  that  if,  in  the  preceding  theorem,  0  is  taken  at  a  vertex, 
the  equation  reduces  to  the  identity  1  =  1. 

Ex.  58.  ABCE  is  any  parallelogram.  AL  is 
any  line  through  A.  Prove  that  the  AAEL  and 
-4i?^are  similar. 

Ex.  59.  From  the  preceding  exercise,  show  that 
in  the  figure  of  that  exercise  the  |  |  EL  •  BK  has 
the  same  value  (that  is,  is  constant),  no  matter  how  AL  is  drawn. 

Ex.  60.  In  the  A  ABC  a  square  is  described 
on  the  altitude  CT.  Let  AL  intersect  CB  in  H. 
If  HF  is  parallel  to  CT  and  HG  is  perpendicular 
to  CT,  prove  that  the  4-side  EGHF  is  a  square. 
(This  square  is  said  to  be  inscribed  in  the 
triangle.) 

Ex.  61.  How,"  then,  would  you  inscribe  a 
square  in  a  given  triangle  ? 

Ex.  62.  If,  in  the  preceding  theorem,  we  had  drawn  BL  in  place  of  AL, 
then  BL  produced  would  in  general  meet  ^C  as  at  G'.  Draw  lines  as 
before,  and  prove  that  the  new  4-side  is  also  a  square.     (Escribed  square.) 

Ex.  63.    How  do  you  construct  an  escribed  square  to  a  triangle  ? 

Ex.  64.  If  a  circle  is  circumscribed  about  a  triangle  whose  two  sides 
are  a  and  6,  and  if  the  altitude  to  side  c  is  h,  and  d  the  diameter  of  the 
circle,  prove  that  a:h::d:b. 

(Join  end  of  diameter  CD  with  A,  and  prove  triangles  similar.) 

Ex  65.  By  multiplying  means  and  extremes  of  the  abovs  proportion, 
what  rectangles  do  you  find  equal  ? 


XV.     SIMILAR   FIGURES  175 

Ex.  66.  If  a  series  of  parallels  cut  any  two  transversals,  the  segments 
determined  by  the  parallels  will  be  proportional.  By  using  this  theorem, 
show  "how  to  divide  a  given  line-segment  AB  into  parts  proportional  to 
any  number  of  given  line-segments,  w,  k,  Z,  etc. 

Of  what  proposition  in  Group  VII  is  the  foregoing  theorem  a  generali- 
zation ? 

Hint.  —U  AT,  CG,  EH,  and  57 be  the  parallels,  and  FI,  AB  the  trans- 
versals, draw  AL  ||  FI.    XV.  1  «,  Sch.  then  gives 

AE^AJ^FH  (VI,  ^ 

CE     JK      GH'  ^  ^ 

whenc>e, 

AE C  £(   EB  /"VV    1  \ 

'FH~~GH~'m'  ^       '    '^ 


XVI.      GROUP   ON  AREAL   RATIOS 

PROPOSITIONS 

XVI.  1.  If  two  triangles  have  an  angle  of  one  equal 
to  an  angle  of  the  other,  they  are  to  each  other  as  the 
rectangles  of  the  sides  respectively  including  the  equal 
angles, 

A 


A  c  L  B        E  g  F 

Hyp.     If,  in  the  A  ABC  and  EFG,  Z.A  =  Z  £7, 

Cone. :  then  A^5(7:  A  EFQ  : :  □  6  •  c  :  □/•  g. 

Dem.     Place  A  EFQ  on  A  ABG  so  that  it  takes  the  position 
AJL.     BmwJB. 

AAJL  lAAJB  .'.AL'.AB.  (1) 

[A  with  equal  altitudes  are  to  each  other,  etc.]  (XIII.  1  c,  Sch.  2.) 

A  AJB  :  A  ABC : :  AJ:  AC.    (Same  reason.)    (2) 

.-.  AAJL :  A  ABC: :  AL  •  AJ:  AB  -  AC 

(Multiplying  (1)  by  (2).) 

/.  AEFG:AABC:icnf'g:C3b-c. 

Or,  by  inversion, 

A  ABC :  A  EFG  ::c3b  •  c:c3f' g. 

Q.E.D 
176 


XVI.     AREAL   RATIOS  177 

XVr.  2.  If  two  triangles  are  similar,  tJmj  are  to  each 
other  as  the  squares  on  any  two  homologous  sides. 


Q 
//        Xe 

A  c  B  E  g"  F 

Hyp.    If  the  triangles  I  and  TI  are  similar, 
Cone.  :  then  Al  :  All  ::  a'  :  e^::  h''  :  f  ::  c^  :  g\ 

Dem.  fjl=l^-  (XVI.1.)(1) 

Now  -  =  -.  (Horn,  sides  of  ^^  A.) 

/      e 

Substituting  in  (1)  for  - ,  its  equal,  -,  we  have 

Al  '  All  ' '  ct^  ■  e^  and  similarly  as  b^  :  f^. 

Q.E.D. 


Ex.  1.   Two  equilateral  A  are  as  5  :  4.     Compare  their  altitudes. 

Ex.  2.  The  base  of  one  equilateral  triangle  equals  the  altitude  of 
another.     What  is  the  ratio  of  their  areas  1 

Ex.  3.  The  homologous  sides  of  two  similar  polygons  are  in  the  ratio 
of  3  :  7.     What  is  the  ratio  of  the  areas  of  the  polygons  ? 

Ex.  4.    What  are  homologous  lines  of  similar  figures  ? 

Ex.  5.   Why  are  the  pefrimeters  of  similar  polygons  homologous  lines  ? 

Ex.  6.  If  two  similar  polygons  have  equal  perimeters,  what  do  you 
know  of  their  areas  ? 

Ex.  7.  Under  what  conditions  are  two  dissimilar  triangles  equal  ? 

Ex.  8.  What  is  the  relation  between  the  areas  of  two  triangles  that 
have  an  angle  of  one  equal  to  an  angle  of  the  other  ? 

Ex.  9.  Draw  a  triangle.  Draw  a  second  triangle  whose  vertex  angle  is 
the  supplemental  adjacent  angle  of  the  other.  Prove  that  the  areas  of 
these  triangles  vary  (or  are  to  each  other)  as  the  rectangles  of  the  sides 
including  the  supplemental  angles. 


178  THE    ELExMENTS   OF   GEOMETRY 

XVI.  3.  If  tioo  polygons  are  sirnilar,  they  are  to  each 
other  as  the  squares  of  any  tivo  homologous  sides. 


A  B  A  B' 

Ryp.     If  polygon  ABC-"  is  similar  to  polygon  ^'^'(7  ••• , 
and  if  their  areas  be  Q  and  Q',  respectively, 

Cone. :  then  Q:Q'::A^:  A^'\ 

Dem.     Draw       ^  FO,  FB,  and  FC,  FB'. 

[Then  the  triangles  of  the  first  polygon  are  similar  to  the 
similarly  placed  triangles  of  the  second  polygon.]         (XV.  5.) 

.     ^^  =  ^  ^^^  ^FS  ^A  BFC  ^FC'  ^  A  FEG 
"  JiB^     AFA'B'     fW     AB'FO      WG^     AFE'C' 

(XVI.  2.) 
.  AjP.^^  ^  A  BFC  ^  AFEC  f  ^  aS  \  ,.  .s 
''  AFA^B'     ABF'O     AFE'C\     JTb'')'       ^   ^'    '^ 


A  FAB  +  BFO  +  FEG  A  FAB       AB" 


(XL  2.) 


AF'A'B'  +  BF'C'  +  FE'G'     AFA'B'     jr^ 
That  is,  Q:Q'::  AB"  :  jVW. 

XVI.  3  a.   TTie  areas  of  tioo  similar  polygons  are  to 

each  other  as  the  squares  of  any  tioo  homologous  lines. 

0em.  AB :  A'B'  r.FBt  F'B'.     (Hom.  sides  ^  A.) 

.-.  Aff :  aW  i'.fS:  FB'-.  (XI.  3,  b.) 

But  Q :  Q'.-.A'b':  AB'\  (XVI.  3.) 

..  QiQ'iiFB'iFB'', 


XVI.     AREAL    RATIOS  179 


XVI.     SUMMARY    OP    PROPOSITIONS    IN    THE    GROUP 
ON    AREAL    RATIOS 

1.  If  two  triangles  have  an  angle  of  one  equal  to  an 
angle  of  the  other,  they  are  to  each  other  as  the  rectan- 
gles of  the  sides  respectively  including  the  equal  angles. 

2.  If  two  triangles  are  similar,  they  are  to  each  other 
as  the  sguares  on  any  two  homologous  sides. 

3.  If  two  polygons  are  similar,  they  are  to  each  other 
as  the  squares  of  any  tivo  homologous  sides. 

a  TJie  areas  of  any  ttvo  similar  polygons  are  to 
each  other  as  the  squares  of  any  two  homolo- 
gous lines. 


180  THE  ELEMENTS  OF  GEOMETRY 

PROBLEMS 

pROB.  I.  On  a  given  line  that  is  to  he  homologous 
to  a  given  side  of  a  given  polygon,  to  construct  a  poly- 
gon similar  to  the  given  polygon. 


Given.  Any 
line  I,  and  any 
polygon  P.  ^ 


Required.  A  polygon  on  I,  ~  P  and  of  which  the  side  equal 
to  I  shall  be  homologous  to  AF. 

Const.  Lay  off  A'F  =  I,  at  A'  construct  ZA'  =  ZA;  at  F' 
construct  Z  F' =  Z  F. 

Find  a  fourth  proportional  to  AF,  A'F',  and  AB. 

Lay  off  this  fourth  proportional  on  the  terminal  line  of  Z  A', 
as  A'B'. 

[Completion  of  construction  and  proof  left  to  student  as  an 
exercise.] 

pROB.  11.  To  construct  a  polygon  similar  to  each 
of  tivo  given  similar  j^olygons  and  equal  to  their 
difference. 

HixT.  —  Construct  a  right  triangle  as  in  Prob.  I,  and  use  XVI.  3. 

pROB.  III.  To  construct  a  polygon  similar  to  each  of 
two  given  polygons  and  equal  to  their  sum. 

Hint.  —  Construct  a  right  triangle  as  in  XIV.  Prob.  I,  and  use  XVI.  3. 

Ex.  10.  Draw  a  triangle.  Draw  a  second  triangle  within  it,  two  of 
whose  sides  are  perpendicular  to  two  sides  of  the  first.  Prove  that  the 
areas  of  these  triangles  vary  as  the  rectangles  of  the  sides  that  are  perpen- 
dicular to  each  other. 


XVI.     AREAL   RATIOS  181 

The  areas  of  two  similar  triangles  are  respectively  196  sq.  ft.  and 
256  sq.  ft. 

Ex.  11.    What  is  the  ratio  of  any  pair  of  their  homologous  sides  ? 

Ex.  12,  What  is  the  ratio  of  the  rectangles  of  the  sides  including  a  pair 
of  homologous  angles  ? 

To  construct  a  triangle  similar  to  ABO  and  satisfying  the  following 
conditions : 

Ex.  13.  Having  a  perimeter  three  times  as  long  as  the  perimeter  of 
ABC. 

,Ex.  14.    Having  an  area  equal  to  four  ninths  the  area  of  ABO. 

Ex.  15.    Having  an  area  twice  as  great  as  that  of  ABO. 

Ex.  16.  The  areas  of  two  triangles,  I  and  II,  having  one  angle  in  com- 
mon, are  20  sq.  ft.  and  60  sq.  ft.,  respectively.  The  sides  in  triangle  I 
about  the  common  angle  are  5  ft.  and  6  ft.  One  of  the  corresponding 
sides  of  triangle  II  is  12  ft.    What  is  the  length  of  the  other  side  ? 

Ex.  17.   The  sides  of  a  triangle  are  4,  9,  10. 

Divide  the  triangle  into  two  equal  parts  in  three  ways  by  drawing,  in 
succession,  parallels  to  the  three  sides. 

G 

/\ 

Ex.  18.   Prove  Theorem  2  by  using  the  adjoining         /  j     \ 
figure,  OH  being  the  altitude  of  the  l^ABO.  e/—^ -\^ 

A     H  B 

Ex.  19.  Each  side  of  a  regular  pentagon  is  3.  Construct  a  similar 
pentagon  twice  as  large  as  the  first. 

Ex.  20.  To  construct  a  hexagon  similar  to  a  given  hexagon  and  having 
one  third  the  area  of  the  given  hexagon. 

Ex.  21.  The  areas  of  two  similar  polygons  are  324  sq.  ft.  and  576  sq.  ft. 
They  are  divided  into  three  sets  of  similar  triangles  by  diagonals  drawn 
from  each  of  two  homologous  vertices. 

What  is  the  ratio  of  the  areas  of  corresponding  pairs  of  the  similar 
triangles  ? 

What  is  the  ratio  of  the  homologous  diagonals  ? 

Ex.  22.  Given  two  similar  hexagons  Q  and  B.  To  construct  a  hexagon 
similar  to  Q  and  B  and  equal  to  their  sum. 

Ex.  23.  Given  two  similar  pentagons  Q  and  R.  To  construct  a  penta- 
gon similar  to  Q  and  B  and  equal  to  their  difference. 


182  TITK    ELEMENTS   OF   GEOMETKY 

Ex.  24.  Ciwn  any  number  of  similar  polygons.     Construct  a  polygon 
similar  to  each  and  equal  to  their  sum. 
(Use  XVI.  a  and  XIV.  Prob.  III.) 

Ex.  25.  To  divide  a  triangle  into  two  equal  parts  by  a  line  parallel  to 
a  given  line. 

Ex.  26.  To  divide  a  triangle  into  two  equal  parts  by  a  line  through  a 
given  point  on  one  side  of  the  triangle. 

Ex.  27.  To  divide  a  triangle  into  three  equal  parts  by  parallels  to  one 
side. 

Ex.  28.  What  does  the  area  of  a  triangle  equal  in  terms  of  the  base 
and  altitude  ? 

Ex.  29.  The  sides  of  a  triangle  are  a,  6,  and  c.  The  altitudes  to  these 
sides  are  h„,  hb,  and  he. 

Show  that  a-  ha  =  b-hi  =  C'  he. 

Ex.  30.   Divide  each  member  of  the  preceding  equation  by  ha  •  h  and 

show  that 

a  _b  _    c 

h     ha     hnhh 

he 

Ex.  31.   Place  ikl^  =  x  and  construct  x  (a  fourth 
he 
proportional  to  fto,  ^j,  and  he). 

Ex.  32.    If  X  equals  a  line  m,  then  —  =  ^-  =  -£. . 

hh     ha     m 

Ex.  33.  Why,  then,  would  a  triangle  whose  sides  are  Aj,  ha,  and  m  be 
similar  to  a  triangle  whose  sides  are  a,  ft,  and  c  ? 

Ex.  34.  Construct  such  a  triangle,  A'B'C,  and  draw  the  altitude 
corresponding  to  he  of  A  ABC.  Produce  this  altitude,  if  necessary,  to 
equal  he.  Through  its  foot,  draw  the  parallel  to  c'.  Produce  the  sides,  if 
necessary,  to  meet  this  parallel.  Prove  this  new  triangle  congruent  with 
A  ABC. 

Ex.  35.   Construct  a  triangle  having  given  the  three  altitudes. 


XVII.     GROUP   ON   LINEAR  APPLICATION 
OF   PROPORTION 

PROPOSITIONS 

XVII.  1.  If  the  bisectors  of  the  interior  and  exterior 
angles  at  the  vertex  of  a  triangle  are  draion,  these 
bisectors  will  divide  the  base  into  segments  proportional 
to  the  other  two  sides. 


Hyp.     If  CT  bisects  AACB  (Fig.  1),  or  /.AGH  (Fig.  2), 

Cone. :  then,  in  either  figure,  AT  :TB::AC:  BC. 

Dem.     Case  I.     Extend  BC,  making  E0=:  AC. 

Draw  EA.     A  CAE  is  isoangular.  (IV.  1.) 

.-.  AACB  =  2Z.E.  (in.  2  a.) 

.-.  Z  BCTf=  ^4^)  =  ^  -27.  (Ax.  2.) 

.-.CTWAE.  (Def.  oflls.) 

.-.  AT:TB::EC:  BC  (XV.  1.) 

.Substituting  AC  for  EC,  we  have 

AT:TB::AO:Ba 

Q.B.P. 

188 


184  THE  ELEMENTS  OF  GEOMETRY 

Dem     Case  II.     Make  EC  =  AG. 

Draw  EA.     A  CAE  is  isoangular.  (IV.  1.) 

.-.  ZACH  =  2  Z  CEA  (III.  2  a.) 

.•    CTII  ^^.  (Det  of  II..) 

.-.  AT'  TB'.'.AC'BC. 

Q.E.D. 

Det.  When  a  line-segment  is  divided  internally  and  exter- 
nally in  the  same  absolute  geometric  ratio,  the  line-segment  is 
said  to  be  divided  harmonically. 

XVII.  3  a  The  bisectors  of  the  interior  and  exterior 
vertex  angles  of  a  triangle  divide  the  base  harmonically 
in  the  ratio  of  the  sides,       jj^ 

flyip.  If  CT  and  CT  bisect 
mt  Z.AOB  and  ext.  Z.  ACH, 
respectively,  and  cut  AB  in  T 
and  T', 

Cone. :  then  AT  -  TB 

Dem  i^=^'y  also  ^^=^~'  (XVII  1.) 

TB     CB  TB     CB 

AT'.TB:    AT  :  T  B.  (Ax.  1.) 

Q.E.D. 

JCVIl.  1  0  If  T  and  T  divide  AB  harmonically,  then 
the  points  a  and  B  divide  the  line  TT  harmonicalty 

For,  if  AT:  TB::Ar:  T'B, 

then,  by  alternation,  AT :  AT'  ii  TB:  T'B. 

That  is,  the  ratio  of  the  distances  of  A  from  T  and  T'  equals 
the  ratio  of  the  distances  of  B  from  T  and  T', 

A^  B  and  T,  T'  are  called  two  pairs  of  conjugate  hctrmorvk 
paints* 


jT 


XVII.    LINEAR   APPLICATION   OF    PROPORTION       185 


XVII.  2    The  perimeters  of  similar  polygons  are  to 
each  other  as  atiy  two  homologous  lines 
Hyp.    If  p  and  p'  be  the 


perimeters  of  two  similar 
polygons,  AB  and  A'B 
two  homologous  sides,  and 
AE  and  A'E'  any  two 
homologous  lines  (e.g.  ho- 
mologous diagonals), 


E 


f> 


Cone. :  then    pip'i.AB.  A'B> . :  AE    A'E 

Pern.         AB :  A'B' :  :  BC  -  B'C  : :  CE  :  C'E',  etc. 

(Horn,  sides  of  ' 
AB  ^BC 
"A'B' 


AB-tBC-hCE-t 


A'B'  -\-B'C'  -\-  C'E'  -I-  •  •      A'B'     B'C 
[In  a  series  of  equal  ratios  the  sum,  etc.  ;] 
i.e.  p:p'::AB:  A'B',  etc. 

But  AE :  A'E'  ..AB:  A'B' 

.'.  p:p':'.AB:  A'B'  ::AE:  A'E' 


w  polygons.) 
etc. 

(XL  2.) 

(XV,  5.) 
(Ax.  1.) 

Q.E.D. 


Ex.  1.  The  sides  of  a  triangle  are  7,  9,  and  12.  Find  the  segments  of 
the  side  12  made  by  the  interior  and  exterior  bisectors  of  the  opposite 
angle. 

Ex.  2.  To  divide  a  line-segment  harmonically  in  a  given  ratio  ;  i.e.  so 
that  the  ratio  of  the  segments  of  internal  and  external  division  shall  equal 
the  ratio  of  two  given  lines  a  and  b. 

Ex.  3.  If,  in  the  last  exercise,  a  =  b,  what  becomes  of  the  external  point 
of  division  of  the  given  line-segment  ? 

Ex.  4.  If  the  diagonal  of  one  pentagon  is  twice  as  long  as  the  corre- 
sponding diagonal  of  a  similar  pentagon, 

How  does  the  perimeter  of  the  first  figure  compare  with  that  of  the 
second  ? 

Ex.  5.  If  the  diagonal  of  a  pentagon  is  equal  to  the  sum  of  the  corre- 
sponding diagonals  of  two  pentagons  similar  to  the  first, 

What  relation  exists  between  the  perimeter  of  the  first  figure  and  the 
perimeters  of  the  other  two  ? 


186 


THE  ELEMENTS  OF  GEOMPrrRY 


XVII.  3.  If  two  angles  have  their  vertices  at  the  cen- 
ters of  two  unequal  circles,  the  angles  are  to  each  other 
as  the  arc  of  the  first  divided  hy  its  radius  is  to  the  arc 
of  the  second  divided  hy  its  radius. 

E 


o  T 

Hyp.     If  Z  BCE  and  Z  K  have  their  vertices  C  and  K  at  the 
centers  of  circles  of  radii  r  and  r',  respectively, 

cone. :  then  Z  JSC£ :  Z /r ::  5]:«^ :  51^. 

r  r 

Dem.     With  C  as  a  center  and  a  radius  =  r\  describe  arci^G^. 

Produce  arc  FG,  making  arc  FH  =  arc  LM. 

Draw  CH. 

Then  ZFCH=ZK. 

[In  equal  CD  equal  A  at  the  centers  intercept,  etc.]      (IX.  3.) 

.-.  ZFCG:ZFCH::2iTcFG:a,vcFH. 

[In  equal  ®  central  A  vary  as  their  intercepted  arcs.]    (XII.  1.) 

But  AFCG  =  Z.BCE. 


ZBOEiZFCH:: 


arc  FG   SiTcFH 


r'  r' 

But  the  sectors  BCE  and  FCG  have  =  central  A. 
.    arci^G^     arc^^ 


(XV.  7.) 


ZBCE-.ZFCH:: 
..  ZBCE:ZK:: 


arc  BE .  arc  FH 

r  r' 

BicBE   SiTcLM 


Q.E.D. 


XVII.     LINEAR  APPLICATION   OF   PROPORTION        187 

ScH.  Upon  this  theorem  is  based  what  is  called  the  Radial 
method  of  measuring  angles. 

The  unit  angle  of  this  system  is  of  course  the  angle  whose  meas- 
ure is  1 ;  i.e.  the  angle  for  which  —  —  1 ;  or  arc  =  radius.    This 

unit  is  called  a  Radian,  and  will  hereafter  be  shown  to  denote 
an  angle  of  about  57°. 3. 

The  advantage  of  radial  measure  over  the  ordinary  measure- 
ment in  degrees  is  that  the  former  does  not  depend  at  all  upon 
the  size  of  the  circle. 

The  radian  is  practically  the  only  unit  employed  in  advanced 
work. 


Application    to    the    Sides    of   a   Right   Triangle   and 
THE   Line-segments   dependent  upon   the  Altitude 

XVII.  4.  Lemma.  If  the  altitude  of  a  right  triangle 
is  drawn  to  the  hypotenuse,  the  three  triangles  of  the 
resulting  figure  are  similar. 

Hyp.  If,  in  rt.  A  ABC, 
the  altitude  CH  is  drawn 
to  the  hypotenuse, 

Cone. :  then     rt.  A I  ~  rt.  A  ABC^  rt.  A II. 

Dem.  Et.  A  .45(7  ~  rt.  A  L 

•.•  Z^  is  common  to  both  of  them. 

[Two  rt.  A  are  ~  if  an  acute  Z  of  one,  etc.]  (XV.  2  a.) 

Similarly,  rt.  A  ABC  ^  rt.  A IL 

/.  the  z^  of  rt.  A I  =  respectively  the  A  of  rt.  A II.     (Ax.  1.) 

.-.  rt.AI'-rt.AIL  (XV.  2.) 

.-.  rt.  AI~rt.A^BO'-rt.  AIL 

OE.D 


188         THE  ELEMENTS  OF  GEOMETRY 

XVII.  5.  Jf  the  altitude  of  a  right  triangle  to  the 
hypotenuse  is  drawn: 

(a)  The  altitude  is  a  mean  proportional  between  the 
segments  of  the  hypotenuse. 

(p)  Either  leg  is  a  mean  proportional  {or  geometric 
mean)  between  the  hypotenuse  and  the  segment  adjacent 
to  that  leg. 

(c)  The  segments  of  the  hypotenuse  are  to  each 
other  as  the  squares  on  the  legs  respectively  adjacent 
to  them. 

(d)  The  square  of  the  length  of  the  hypotenuse  equals 
the  sum  of  the  squares  of  the  lengths  of  the  legs. 

Hyp.    If  CL  is  ^ 

the  altitude  to  AB^ 
the  hypotenuse  of 
the  rt.  A  ABC, 

Cone. :  then 

(a)  ALiCLi.CL:  LB,  or  CZ' =  AL  -  LB. 

(b)  ALiCAi.CA:  AB,  or  CA"  =  AL  -  AB. 

(c)  AC'iBC'-.'.ALiLB. 

(d)  ac'  +  bc'^ab'. 

Dem.     (a)  Tit  A  ALC  <^  vt  A  CLB.  (XVII.  4.) 

.-.  AL:CL::CL:  LB.     (Horn,  sides  of  ^  A,  etc.) 

..  CL'  =  AL'LB.  (XL  1.) 

Q.E.D. 

Dem.     (6)  m.  A  4LC '^  rt.  A  ACB.  (XVII.  4.) 

.-.  AL:CA::CA:AB,  or  CA'  =  AL'AB',  (1) 

also,  LB  :BC::BC:  AB,  ot  BC"  =  LB  -  AB.  (2) 

Q.E.D. 


XVII.    LINEAR  APPLICATION  OF  PROPORTION        189 

Dem.     (c)  Divide  (1)  by  (2),  member  by  member,  and  we  have 

AG" :  BC^  iiAL:  LB. 

Q.E.D. 

Dem.     (d)  Add  (1)  to  (2),  member  to  member,  and  we  have 

AG"  -hB(f=  (AL  +  LB)  •  AB 

=  AB'AB 

=  aS. 

Q.E.D. 

Def.     If  0  be  any  point  on  the  straight  line  AB  (whether 

between  the  points  A  and  B,  or  on  AB  produced),  the  distances 

OA  and  OB  are  called  the  segments  of  AB  made  by  the  point  0. 

Exe  6.    What  are  homologous  sides  of  similar  triangles  ? 

Name  the  sets  of  similar  triangles  in  the  figure  of  XVII,  4. 

Read  the  homologous  angles  of  each  set ;  also  the  homologous  sides. 

Ex.  7.    Find  a  mean  proportional  to  a  and  h  by  using  XVII.  6  a. 

Ex.  8.    Find  a  third  proportional  to  a  and  6  by  using  XVII.  6  a. 


Ex.  9.    Given  a  =  8  ft.,  c  =  13  ft.,  and 
s=z\\lt.     Find/. 


c 
Ex.  10.    Given  :  perimeter  of  small  triangle  is  46  ft.  ;  of  large,  138  ft. 
What  is  the  ratio  of  any  two  homologous  sides  ? 

Ex.  11.  In  two  similar  triangles  the  sides  of  the  first  are  4  ft.,  9  ft.,  and 
11  ft. 

The  shortest  side  of  the  second  is  12  ft. 
What  is  the  length  of  the  other  two  sides  •? 
What  is  the  ratio  of  their  perimeters  ? 

The  sides  of  a  triangle  are  7  ft.,  10  ft.,  and  12  ft.     Find : 

Ex.  12.  The  segments  determined  on  each  side  by  the  bisector  of  the 
interior  angle  at  the  opposite  vertex. 

Ex.  13.  The  segments  determined  on  each  side  by  the  bisector  of  the 
exterior  angle  at  the  opposite  vertex. 

Ex.  14.   The  projections  of  each  side  on  the  other  two. 

Ex.  15.  The  projection  of  the  bisector  of  each  interior  angle  on  the 
opposite  side. 


190 


THE  ELEMENTS  OF  GEOMETRY 


XVII.  6.  If  through  a  fixed  point  any  line  is  drawn 
cutting  a  circle  in  two  points,  the  rectangle  of  the  seg- 
ments of  this  line  is  constant,  in  whatever  direction  the 
line  is  drawn. 

Three  Cases. 

(1)  T'he  fixed  point  within  the  circle. 

(2)  The  fixed  point  on  the  circumference, 

(3)  The  fixed  point  loithout  the  circle. 


0 
Casb  (1). 


Case  (3). 


Hyp.     If  0  is  a  fixed  point,  and  AB  is  any  line  through  0, 
arid  EG  is  any  other  line  through  0,  each  cutting  the  circle, 

Cone. :  then  the  □  of  J50  •  0^  =  the  □  of  ^0  •  OC. 

Dem.    Draw  AE  and  BC  in  each  of  the  three  cases. 

AAOE-^ABOO. 

'.'  ZA  =  ZC.    ('Each  is  measured  by  ^^^^^.\ 

ZAOE  =  ZBOa     (Why?) 

.-.  A  AOE  ~  A  BOC,  and  OA:OE::00:  OB.   (XV.  2.) 

.%  the  rectangle  of  BO  -  0A  =  the  rectangle  of  EO  •  00. 

Q.E.D 

ScH.     The  proportion  OA  :  OE : :  OC:  OB  may  be  written 

OA^qc        O^^i  .  OB 
OE     OB'        OE         '  OC 


XVII.     LINEAR  APPLICATION  OF  PROPORTION        191 

That  is :  The  ratio  of  two  segments  equals  the  reciprocal  of 
the  ratio  of  the  two  corresponding  segments ;  in  other  words, 
the  segments  are  reciprocally  proportional. 

XVII.  &  a.  If  from  the  same  point  to  the  same  circle 
a  tangent  and  a  secant  he  drawn,  the  tangent  ivill  he  a 
mean  proportional  hetween  the  secant  and  the  part 
without  .the  circle.  ^ . 

Hyp.     If  from  0,0T  f  \}^^ 

a    tangent,    and    OB   a  I  ^^ ^.^^-^n/ 

secant    are    drawn    to  \  ^...^^^^    jl^ 

Cone. :  then  OB.OT.iOT:  OA,  or  0T'  =  0B  •  OA. 

Dem.     Draw  AT  and  BT. 

A  OAT r^  A  0TB.  (XV.  2.) 

.-.  OB:OT::OT:  OA,  or  OT' =  OB  -  OA        (XL  1.) 

Q.E.D. 
ScH.     OA  is  a  tliird  proportional  to  OB  and  OT. 

OB  is  a  third  proportional  to  OA  and  OT. 

The  sides  of  a  triangle  are  7  ft.,  10  ft.,  and  12  ft.     Find  : 
Ex.  16.    The  projection  of  the  bisector  of  one  of  the  exterior  angles  on 
the  opposite  side. 

Ex.  17.    The  lengths  of  any  of  the  angle  bisectors. 

Ex.  18.   The  area  of  the  triangle. 

Ex.  19.  The  acute  angles  of  a  right  triangle  are  60°  and  30°.  What  is 
the  ratio  of  the  segments  into  which  the  shortest  side  is  divided  by  the 
bisector  of  the  opposite  angle  ? 

Ex.  20.  Given  a  circle  of  radius  a  ;  through  a  point  at  a  distance  2  a 
from  the  center,  tangents  are  drawn.     Find : 

(a)  The  length  of  each  tangent.     (6)  The  length  of  the  chord  of  contact. 

(c)  The  angle  between  the  tangents. 

(d)  The  an^le  at  the  center  subtended  by  the  chord  of  contact. 

(e)  The  distance  of  the  chord  of  contact  from  the  center. 


192  THE    ELEMENTS   OF   GEOMETRY 

XVII.     SUMMARY    OF   PROPOSITIONS    IN    THE   GROUP 
ON   THE  LINEAR  APPLICATION   OF   PROPORTION 

Application  to  Angle  Bisectors,  Perimeters, 
AND  TO  Angular  Measurement 

1.  If  the  bisector  of  the  interior  and  exterior  angles 
at  the  vertex  of  a  triangle  are  drawn,  these  bisectors 
will  divide  the  base  into  segments  proportional  to  the 
other  two  sides, 

a  The  bisectors  of  the  interior  and  exterior  vertex 
angle  of  a  triangle  divide  the  base  harmoni- 
cally in  the  ratio  of  the  sides. 

2.  The  perimeters  of  similar  polygons  are  to  each 
other  as  any  two  homologous  lines. 

3.  If  two  angles  have  their  vertices  at  the  centers  of 
two  unequal  ciixles,  the  angles  are  to  each  other  as  the 
arc  of  the  first  divided  by  its  radius  is  to  the  arc  of  the 
second  divided  by  its  radius. 

ScH.     Unit  of  angular  magnitude :  radian. , 

Application  to  the  Sides  of  a  Eight  Triangle  and  the 
Line-segments  dependent  upon  the  Altitude 

4.  Lemma.  If  an  altitude  of  a  light  triangle  is  drawn 
to  the  hypotenuse,  the  thtee  triangles  of  the  resulting 
figure  will  be  similar. 

5.  If  the  altitude  of  a  right  triangle  to  the  hypotenuse 
is  drawn, 


XVIL     LINEAR    APPLICATION    OF    PROPORTION        198 

(a)  The  altitude  is  a  mean  proportional  between 

the  segments  of  the  hypotenuse. 

(b)  Either  leg  is  a  mean  proportional  (or  geometric 

mean)  hetiveen  the  hypotenuse  and  the  segment 
adjacent  to  that  leg. 

(c)  The  segments  of  the  hypotenuse  are  to  each 

other  as  the  squares  on  the  legs  respectively 
adjacent  to  them. 

(d)  The  square  of  the  length  of  the  hypotenuse 

equals  the  sum  of  the  squares  of  the  lengths 
of  the  legs. 

Application  to  Chords,  Tangents,  and  Secants 

6.  If  through  a  fixed  point  any  line  is  drawn  cutting 
a  circle  in  two  points,  the  rectangle  of  the  segments  of 
this  line  is  constant,  m  whatever  direction  the  line  is 
drawn. 

Three  Gases. 

(1)  The  fixed  point  within  the  circle. 

(2)  The  fixed  point  on  the  circumference. 

(3)  The  fixed  point  loithout  the  circle. 

ScH.  The  ratio  of  two  corresponding  segments  equals  the 
reciprocal  of  the  ratio  of  the  other  two  corresponding  segments. 
That  is,  the  segments  are  reciprocally  proportional. 

a.  If  from  the  same  point  to  the  same  circle  a  tan- 
gent and  a  secant  are  drawn,  the  tangent  ivUl 
he  a  mean  proportional  hetiveen  the  secant 
and  the  part  without  the  circle, 

ScH.     On  third  proportionals. 


194 


THE  ELEMENTS  OF  GEOMETRY 


PROBLEMS 

Prob.  I.     To  find  a  fourth  proportional  to  three  given 
l^nes.  a    b   q   X 

Given.     The  three  lines,  a,  b,  and  c. 

Required.     A  fourth  proportional  to  a,  6, 
and  c. 

Analysis.     Suppose  x  to  be  the  required  fourth  proportional. 
Then  a:b::c:x. 

Suggested  Theorem,  XVII.  6,  Cases  (1)  and  (3). 


Fio.  1. 
The  "  dash  and  dot "  Hue  in  figures  is  the  line  required. 

Const.     Case  (1).   Draw  a  circle  whose  diameter   >  6  -f  c. 

(Fig.  1.) 

In  this  circle  draw  a  chord  BC  =  b  +  c. 
Take  OA  =  a  and  produce  it  to  intersect  circle  at  X. 
Then  OX  is  the  required  fourth  proportional.    (XVII.  6  (1).) 

Q.E.F. 

Const.     Case  (2).   Draw  a  circle  whose  diameter  >  6  —  c. 
(Big.  2.) 

In  this  circle  draw  a  chord  B0=  b  —  c, 
.    Produce  it  so  that  CO  =  c. 

Take  OA  =  a. 

Then  OX  is  the  required  fourth  proportional.     (XVII.  6  (3).) 

...Q.E.F. 


XVII.     LINEAR  APPLICATION   OF    PROPORTION        195 

Note.  —  The  student  should  suggest  still  other  ways  of  laying  off  the 
given  segments  in  Problem  I.  a;  is  a  fourth  proportional  to  a,  6,  and  c 
in  any  one  of  the  following  arrangements  : 

(1)  ff  :  6  :  :  c  :  X.  (3)  a  :  a;  : :  6  :  c. 

(2)  a  :  &  : :  X  :  c.  (4)  x  :  6  : :  a  :  C. 
Also  in  their  transformations. 

ScH.  If  two  of  the  given  line-segments  are  equal,  e.g.  if  6  =  c, 
each  of  the  constructions  given  above  will  determine  a  third 
proportional  to  a  and  h,  a  and  c,  or  h  and  c,  provided  that  the 
line-segments  that  become  equal  are  not  in  the  same  couplet  of 
the  proportion. 

Prob.  II.    To  construct  a  mean  proportional  to  two 

given  lines. 

Given.      The   lines 
a  and  h. 

Required.     To  con- 
struct a  mean  propor-  ^     ^  ^        t.'^   o 
f  ^   ^                                   Fig.  1.  Fig.  2. 
tional  to  a  and  o. 

Const.     Case  (1).   Draw  EF=  h  and  lay  off  EQ  =  a. 
Describe  a  semicircle  on  EF  as  a  diameter. 
At  Q  erect  a  perpendicular  to  EF  intersecting  semicircle  in  T. 
ET  is  the  required  mean  proportional. 

Proof.         EF:  ET: :  ET:  EQ  ox  h  :  ET: :  ET:  a. 
[Either  leg  of  a  rt.  A  is  a  mean  proportional,  etc.]     (XVII.  5  h.) 
.-.  ET  of  Fig.  1  is  a  mean  proportional  to  a  and  b. 

Const.     Case  (2).   Draw  EQ  =  a  and  EF=  b. 

Then  QF=a-{-b. 

On  QF  as  a  diameter  describe  a  semicircle. 

At  E  erect  a  ±  to  QF  intersecting  the  semicircle  at  T 

ET  is  the  required  mean  proportional. 

Proof.         QE:ET::ET:  EF  or  a  :  ET: :  ET:  b. 

For  QTF  is  2i  right  A.     (Why  ?) 

.-.  ET  of  Fig.  2  is  a  mean  proportional  to  a  and  b. 


196         THE  ELEMENTS  OF  GEOMETRY 

Prob.  III.  To  draw  a  tangent  of  given  length  to  a 
circle  so  that  the  chord  of  the  secant  drawn  from  the 
outer  extremity  of  the  tangent  to  the  same  circle  shall 
equal  a  given  line. 

Given.  The  leogth 
q,  the  line  d,  and  the 
circle  K. 

Required.  To  draw  a  tangent  =  q,  so  that  the  chord  of  the 
secant  from  the  outer  extremity  of  the  tangent  =  d. 

Analysis.     Suppose  tangent  AT=  q  and  that  AO  —  AB  =  d. 

It  follows  that  if  /Of  be  drawn  _L  BC,  M  is  the  mid-point 
oiBC.     (Why?) 

Propositions,  etc.,  suggested  are  : 

Find  the  locus  of  the  mid-points  of  all  chords  in  the  given 
circle  equal  to  (7,  and 

Draw  to  the  given  circle  a  tangent  equal  to  q,  and 

From  the  outer  extremity  of  this  tangent  draw  a  tangent  to 
the  locus  circle. 

We  have  now  discovered  a  method  of  constructing  the  line 
required. 

Discussion.  As  two  tangents  may  be  drawn  from  A  to  the 
locus  circle,  there  are  evidently  two  solutions. 

These  solutions  become  coincident '  when  d  equals  the 
diameter  of  the  given  circle. 

There  is  no  solution  when  d  is  greater  than  the  diameter  of 
the  given  circle. 

Dei  A  line  is  divided  in  extreme  and  mean  ratio  (or  in 
golden  section)  when  one  segment  of  the  line  is  a  mean 
proportional  between  the  whole  line  and  the  other  segment. 


Ex.  21.    The  line  of  centers  of  two  circles  is  divided  externally  in  the  ratio 
of  their  radii  by  its  point  of  intersection  with  an  external  common  tangent. 


XVII.     LINEAR   APPLICATION   OF  PROPORTION        197 


Prob.  IV.    To  divide  a  given  line  in  extreme  and  mean 
yxtio. 


AB 


Q' 
Given.     The  line  AB. 

Required.     To  divide  A'B  in  extreme  and  mean  ratio. 

Const.     Draw  KB  ±  AB.    Let  it  equal  or  be  greater  than 

From  A  draw  a  secant  AM  to  this  circle  so  that  the  chord 

LM=AB.  (Prob.  III.) 

With  ^  as  a  center,  and  a  radius  equal  to  AL,  describe  an 

arc  cutting  AB,  as  at  Q. 

Then  AB  is  divided  at  Q  in  extreme  and  mean  ratio. 

That  is  AB '.AQ'.:AQ:  QB. 

Q.E.F. 
Proof.     AB  is  tangent  to  large  circle  K. 

[If  a  line  is  ±  to  a  radius  at  its  extremity,  etc.]       (IX.  4  a.) 
.-.  AMiAB::  AB.AL.  (1) 

.\  AM-AB'.AB::AB-AL:AL. 

(From  (1)  by  division.) 
But  AB  =  LM, 

and  AL^AQ.  (Const.) 

.-.  AM-  AB  =  AM-  LM  =  AL, 
and  AB-AL  =  AB-AQ=QB. 

.-.  AQ.AB::  QB:AQ.  (2) 

.-.  AB:AQ::AQ:  QB. 

(From  (2)  by  inversion.) 

Q.E.D. 

Discussion.     With  ^  as  a  center,  and  a  radius  equal  to  AM, 
describe  an  arc,  cutting  AB  produced  to  the  left,  as  at  Q'. 
Then  AB:AQ'::AQ':Q'B. 

Prove  by  taking  (1)  by  composition. 
.-.  Q'  is  a  second  point  of  golden  section. 


198         THE  ELEMENTS  OF  GEOMETRY 

Prob.  V.    To  construct  a  square  that  shall  he  three 
times  a  given  square.  ^"-----J 

Given.     The  square  A-C.  /  j  ^\ 

Required.    To  construct  a  A[ B  \      \ 

J r  ----i----f---  - 

square  equal  to  three  times  \     ^**a 

the  square  A-C.  J /7---J — J 

Const.     Produce  ABy  making  AO  =  3  AB. 
Produce  AO,  making  a'  =  a. 

On  AH  as  a  diameter,  describe  a  semicircle. 
At  O  erect  a  perpendicular  to  AH,  meeting  the  semicircle 
in  J. 

JG  is  the  side  of  the  required  square. 

Proof  is  left  to  the  pupil. 

Def.     To  transform  a  figure  means  to  change  it  to  another 
figure  that  is  eqiial  to  the  first. 

Prob.  VI.     To  transform  a  rectangle  into  a  square, 

^F 

Given.     Then  ^-5. 


^Q 


Required.  To  trans- 
form it  into  an  equal 
square.  

Const.     Produce  EC,  making  a'  =  a.    . 
On  EG  as  a  diameter  describe  a  semicircle. 
At  C  erect  a  perpendicular  to  EG,  meeting  the  semicircle 
inF. 

CF  is  the  side  of  the  required  square. 

Proof  is  left  to  the  pupil. 

Ex.  22.  The  line  of  centers  of  two  circles  is  divided  internally  in  the  ratio 
of  their  radii  by  its  point  of  intersection  with  an  internal  common  tangent. 

Ex.  23.  The  points  of  intersection  of  the  pairs  of  internal  and  external 
common  tangents  to  two  circles  divide  the  line  of  centers  harmonically. 


XVII.     LINEAR   APPLICATION   OF   PROPORTION        199 

Pkob.  VII.     To  transform  a  triangle  into  a  square. 

Construct  a  mean  proi^ortional  between  the  base  and  half  the 
altitude. 

Ex.  24.    Two  sides  of  a  right  triangle  are  20  and  21. 

Determine :  (a)  The  projections  of  these  sides  on  the  hypotenuse. 
(6)  The  altitude  on  the  hypotenuse. 

Ex.  25.  If  the  segments  of  a  diameter  be  3  and  7,  what  is  the  length  of 
the  perpendicular  to  the  diameter  at  the  point  of  division  and  extending 
to  the  circumference  ? 

Ex.  26.    Show  how  to  construct  \/l4  geometrically. 

Ex.  27.  What  is  the  expression  for  the  rectangle  of  the  segments  of  a 
diameter  determined  by  any  point  in  the  diameter  ? 

From  this  expression  show  that  the  rectangle  of  the  segments  of  a 
diameter  is  greatest  when  the  segments  are  equal. 

Ex.  28.  If  two  chords  be  drawn  from  any  point  of  a  semicircle  to  the 
extremities  of  a  diameter,  the  squares  of  the  chords  will  be  to  each  other 
as  the  projections  of  the  chords  on  the  diameter. 

Ex.  29.   The  area  of  a  triangle  equals  the  rectangle  of  its  base  by  half 

the  altitude  or  6  .  -• 

2 

Ex.  30.    Construct  a  square  equal  to  twice  a  given  triangle. 
Ex.  31.    Construct  a  square  equal  to  the  sum  of  two  given  triangles. 
Ex.  32.   Construct  a  square  equal  to  the  sum  of  a  given  triangle  and  a 
given  parallelogram. 

Ex.  33.  Construct  a  square  equal  to  the  sum  of  a  given  pentagon  and  a 
given  rectangle. 

Ex.  34.  What  is  the  length  of  the  side  of  a  square  equal  to  an  equi- 
lateral triangle  each  side  of  which  is  10  ft.  ? 

What  is  the  geometric  meaning  of  the  following  expressions? 


Ex.  35. 

c             h 

Ex.  36. 

x=^AB^',  x=^IAB\ 

Ex.  37. 

a;  =  V2  AB^  -  □  of  CE  -  HJ. 

Ex.  38. 

X  =  \/2  J  of  a  •  6. 

Ex.  39. 

^Mq^-'iAw' 

200 


THE   ELEMENTS  OF  GEOMETRY 


E 


Prob.  VIII.  To  construct  a  rectangle  whose  perimeter 
is  equal  to  twice  the  length  of  a  given  line  and  whose 
area  equals  that  of  a  given  square^ 

Given.  The  line 
/,  and  the  square  IS 

Required.  To  con 
struct  a  rectangle 
whose  perimeter 
equals  2 1  and  whose 
i^j-ea  equals  «S 

Analysis.     Suppose  CD  R  has  perimeter  =  2  ^  and  area  =  C  S. 
That-is,  that  rectangle  of  XY-  XZ  =  AB^,  (1) 

and  XY-{'XZ=l.  (2) 

It  follows  from  (1)  and  (2)  that  AB  is  a  mean  proportional 
between  two  line-segments  whose  sum  =  I. 

Several  previously  established  theorems  are  suggested  by  the 
last  statement,  viz. :  XVII.  5  (a)  and  5  (5) ;  XVII.  6  and  6  a ; 
Problems  I,  II,  III.  Of  these,  let  us  try  XVII.  5  a  and 
XVII.  6. 


Case  1. 


C4.se  2. 


Const    Case  1.   Describe  a  seipicircle  on  EF—  I  as  diameter. 

Let  m  be  the  locus  of  a  point  that  is  AB  distance  from  EF. 

From  (7,  the  intersection  of  this  locus  with  the  circle,  draw 

CQ±EF. 

The  rectangle  of  EQ  •  QF  is  the  rectangle  required. 

Q.E.P 


XVII.    LINEAR   APPLICATION    OF   PROPORTION       201 

Proof.     CQ  =  AB,  •■■  m  is  the  locus  of  all  points  AB  distant 
from  EF  ^^^,  ^^.^^^^  ^ 

EQCQ:^  CQ .  QF  or  CQ'  =  rectangle  of  EQ  -  QF. 

(.XVI1.5a) 
QE.D 
Const.     (Jase  2    Draw  any  circle  whose  diameter  >  I 

Draw  a  chord  CJ=  2  AB. 

Draw  the  circle  that  is  the  locus  of  mid-points  of  chords  —I. 

From   Q,  the  mid-point  of  CJ,  draw  a  chord  of  the  larger 

circle  that  is  tangent  to  the  locus  circle. 

The  rectangle  of  EQ  •  QF  is  the  rectangle  required 

QE.P 
Proof  to  be  given  by  the  pupil  from  XVII.  6. 


Ex  40.  If  a  tangent  to  a  circle  be  terminated  by  two  parallel  tangents, 
the  rectangle  of  the  segments  of  this  tangent  determined  by  the  point  of 
tangency  equals  the  square  on  the  radius. 

Ex..  41,  Given  a  circle  of  radius  20  ft. ;  through  a  point  16  ft.  from  the 
center  a  chord  is  drawn. 

Find  the  rectangle  of  the  segments  into  which  the  chord  is  divided  at 
the  point.   • 

Ex.  42.    Given  a  circle  of  radius  24  ft. ;  through  a  point  40  ft.  from  the 
center  a  tangent  is  drawn,  and  also  a  secant  58  ft.  long.     Find  : 
(a)  The  length  of  the  tangent. 
(6)  The  length  of  the  chord  cut  from  the  secant, 
(c)   The  distance  of  the  secant  from  the  center. 

Ex.  43.  The  radius  of  a  circle  is  12  ft. ;  tangents  are  drawn  to  this 
circle  through  a  point  20  ft.  from  the  center.  What  is  the  length  of  the 
chord  joining  the  points  of  tangency  ? 

Ex.  44.    Given  a  line  m  and  a  parallelogram  of  base  b  and  altitude  h. 

Use  Theorem  6  to  construct  on  m  as  a  base  a  rectangle  equal  in  area  to 
the  given  parallelogram. 

Ex.  45.  To  pass  a  circle  through  two  given  points  and  tangent  to  a 
given  line.     (Use  XVII.  6  a.) 

Ex.  46.   To  transform  a  parallelogram  into  an  equal  square. 

Ex.  47.  To  transform  a  scalene  triangle  into  an  equilateral  triangle  of 
the  same  area. 


202 


THE  ELEMENTS  OF  GEOMETRY 


Prob.  IX.    To  construct  a  square  that  shall  he  to  a 
given  square  in  a  given  ratio. 


,^i^        ^^N 

k 

a 

/ 

\    \ 

'/i^  p 

^^^1 

jj 

Am     B  k  € 

Given.     The  square  on  a  and  the  ratio  k  :  m. 
Required.    To  construct  a  square  that  shall  be  to  a*  as  A;  is  to  m. 

Const.     On  indefinite  line  AL,  lay  off  AB  =  m  and  BC=k. 

On  AC  as  a  diameter  describe  a  semicircle. 

At  B  erect  a  perpendicular  to  AC,  cutting  semicircle  in  0- 

Draw  OA  and  OC  and  lay  off  OQ  =  a. 

Draw  GE  II  AC. 

OE  is  the  side  of  the  required  square. 

Q.E.P. 

Proof.    A  FOQ  ^  A  BOA  and  A  FOE  '^  A  BOO.      (XV.  2.) 


and 


But 


or 


.-.  m :  GF 

k'.FE 

,'.  m :  GF 

.:  m:k 

OF'.FE 

m:  k 


:BO:FO  (Hom.  sides  of  ^A.) 
:  BO  :  FO.  (Hom.  sides  of  ^A.) 
:k:FE  (1).  (Ax.  1.) 

:  GF:  FE.  (Taking  (1)  by  alt.) 
:  OG'  :  O^,  (XVII.  5  (c).) 


:  a^ :  OE' 


QJSJ>. 


Prob.  X.    To  construct  a  polygon  similar  to  a  given 
polygon  and  having  a  gioen  ratio  to  it. 


Given.  The  polygon 
P,  and  the  ratio  k :  m. 

Required.  To  con- 
struct a  polygon  similar 
to  P  and  having  with  P 
the  ratio  k :  m. 


XVII.     LINEAR  APPLICATION   OF  PROPORTION       203 

Analysis.  Suppose  X  ^  P  and  X:  P::k:m. 

If  X^P, 

then  X:P::YZ':Aff;                      (XVI.  3.) 

and,  if  X:  P::k:m, 

then  k:m::YZ':  Iff.                         (Ax.  1.) 

This  proportion  suggests  XVII.  5  (c)  and  the  preceding 
problem. 

Hence,  the  construction. 

Const.  Construct  a  line  A'B'  such  that  the  square  upon  it 
shall  be  to  the  square  on  AB  as  A; :  m.  (Prob.  IX.) 

On  this  line  AB',  homologous  to  AB,  construct  a  polygon 

Q  similar  to  P. 

Q  is  the  polygon  required. 

Q.E.F. 
Proof.     (Let  pupil  give  proof.) 


Ex.  48.  To  transform  a  scalene  triangle  into  an  equal  isosceles  triangle 
with  a  given  base  angle. 

Ex.  49.  To  construct  a  square  that  shall  be  to  a  given  pentagon  as  4 
is  to  9. 

Ex.  50.  How  does  an  angle  inscribed  in  a  segment  compare  with  the 
angle  between  the  base  of  the  segment  and  a  tangent  at  either  extremity 
of  the  base  ?    State  the  converse  of  the  proposition. 

Ex.  51.  A  circle  is  described  on  a  given  line  AH  as  a  diameter.  A£ 
is  a  fixed  chord.  BF  is  drawn  perpendicular  to  AH.  AJ is  any  chord 
cutting  BF  in  3. 

Why  is  Z  AJB  =  Z  ABF  ? 

Ex.  52.  Why,  then,  is  AB  tangent  to  a  circle 
through  E,  B,  and  J? 

Prove,  then,  that  AJ?  =  rectangle  of  AE  •  A  J. 

Ex.  53.  (a)  Show  that  the  4-sideC^^J  is  cyclic. 

(&)  Noting  that  ZABH  is  a  right  angle,  and 
that  BC  is  the  altitude  on  the  hypotenuse,  give 
another  proof  that  AB^  =  rectangle  of  AE  •  AJ. 

Ex.  54.  What  is  the  locus  of  a  point  E  that  divides  all  chords  through 
A  so  that  EA  is  a  third  proportional  to  AB  and  the  whole  chord,  say  AJ? 


204         THE  ELEMENTS  OF  GEOMETRY 

Pros.  XI.  To  construct  a  polygon  whose  sides  shall 
be  in  a  given  ratio  to  the  sides  of  a  given  polygon, 

E 

Given.     Any  rj-gon  _1'_ 

u4BC'--- and  two  lines  ^''^^S'-^^J^^-.y^^^o     /^      \ 

k  and  w,  representing  ""~""*"^\-7r  y^\   ^ 

any  given  ratio.  m i ''••■r'^.*"~-\  ^  / 

Required.  Ann-gon  -i  -^ 

~  LMO"-  whose  sides  are  to  the  sides  of  n-gonABC"'  in  the 
ratio  of  k  to  m. 

That  is,  the  ratio  of  similitude  of  the  required  to  the  given 
n-gon  is  A: :  m. 

Const.  Find  a  fourth  proportional  to  k,  m,  and  any  side  AB 
of  n-gon  ABC  '".  (v.  Prob.  I.) 

Let  this  line  be  LM,  and  draw  it  parallel  to  AB  at  any  con- 
venient distance  from  AB. 

Draw  AL  and  BM,  and  let  them  intersect  at  K. 

Draw  KC,  KE,  and  KF. 

Through  L  draw  LQ  II  AF  and  terminating  on  KF, 

Similarly,  draw  QP  and  PO.  Draw  OM, 

The  w-gon  LMO  '"is  the  n-gon  required. 

Q.E.P. 

Proof.  LM:  AB  ::k:m  (Const.) 

KM:KB::KL:KA::LQ:AF,  etc. 
KOiKO. 
.-.  OM  II  BC. 

[If  a  line  divides  two  sides  of  a  A  proportionally,  etc.  ]  (XV.  1  a.) 

.-.  n-gon  LMO n-gon  ABC  —  (XV.  6  a.) 

and  its  ratio  of  similitude  to  n-gon  ABC  •••  is  A; :  m. 

Q.E.D. 

Ex.  55.  What  is  locus  of  J  (Fig.  for  Ex.  53)  so  taken  that  the  rectangle 
of  AE  •  AJ  =  AB^,  where  ^  is  a  fixed  point  in  the  mid-perpendicular  to 
the  given  line-segment  FB  ? 


-fflCki. 


XVII.     LINEAR   APPLICATION   OF   PROPORTION       205 

pROB.  XII.  To  construct  a  rectangle  ivhose  area  shall 
equal  a  given  square,  and  the  difference  of  ivhose  base 
and  altitude  shall  equal  a  given  line. 

Given.     The  square  Q 
and  the  line  d. 


Required.  To  con- 
struct a  rectangle  whose 
area  shall  equal  Q,  and 
the  difference  of  whose  base  and  altitude  shall  equal  d. 


Analysis.     Suppose  [Z}A-L=  the  rectangle  required. 

That  is,  that  □  A-L  =  D  Q  and  that  x  —  y  =  d. 

If  □  A-L  =DQ,  then  X'y  =  q\     (1)  (XIII.  2.) 

Equation  (1)  suggests  any  one  of  the  theorems  concerning 
mean  proportionals,  the  most  convenient  of  which  for  practical 
purposes  is  that  on  the  tangent  and  secant. 

Again,  equation  (1)  is  indeterminate ;  but,  if  we  examine  it 
and  remember  that  x  —  y  =  d,  we  see  that  the  problem  may  now 
be  stated:  Draw  a  tangent  of  a  given  length  (q)  to  a  circle 
so  that  the  chord  of  the  secant  drawn  from  the  extremity  of 
the  tangent  to  the  same  circle  shall  equal  a  given  line  (d). 

(XVII.  Prob.  III.) 

We  have,  therefore,  discovered  or  rediscovered  the  method  of 
constructing  the  required  rectangle.  Por  the  tangent  equals  the 
side  of  the  square,  the  whole  secant  equals  the  base,  and  its  ex- 
ternal segment  equals  the  altitude,  of  the  required  rectangle. 


Ex.  56.  If  AH  is  the  diameter  of  a  circle,  and 
CE  is  perpendicular  to  AH  produced,  prove  that 
the  locus  of  a  point  J  which  divides  any  line  from 
A  to  BF  so  that  the  rectangle  of  AE  •  AJ=  rectangle 
of  AH'  AC,  is  the  circle  on  AH SiS  a  diameter. 

(Note  that  4-side  HCJE  is  cyclic.) 


E   B 


206  THE  ELEMENTS   OF   GEOMETRY 

pROB.  XIII.  To  construct  a  polygon  similar  to  one 
given  polygon  and  equal  in  area  to  another  given  polygon. 

Given.       The  / ^^^^^ 

polygons  P  an  (IQ.        ^^\       /  ^-^^         ^/ 

Required.       lo      \      "^     /         \  ^        \       X       I 

construct  a  poly-        \ I  \ ^^  \  / 

•    '^      \     r»        AH  \ L 

gon  similar  to  F  ^'        b 

and  equal  in  area  to  Q. 

Analysis.     Suppose  polygon  X  ~  P  and  =  Q. 

If  X -  P,  then        P.XiiAff:  A^\  (XVI.  3.) 

In  this  proportion  substitute  for  X  its  equal  Q,  and  we  have 

PiQiiAffiA^. 
This  proportion  contains  but  one  unknown  quantity,  namely, 

The  question  arises.  How  shall  we  construct,  geometrically. 

Now  P  and  Q  are  dissimilar  polygons  by  hypothesis ;  but,  if 
we  change  each  to  an  equal  square  whose  sides  are  respectivelv 
m  and  k,  proportion  (1)  becomes 

.-.  m:k::AB:A'B'. 
That  is,  A'B'  is  a  fourth  proportional  to  m,  k,  and  AB. 
Hence,  the  construction. 

Const.    Change  P  to  an  equal  triangle ;  also  Q.  (XIII.  Prob.  I.) 

Find  a  square  equal  to  each  of  these  triangles. 

(XVII.  Prob.  VII.) 

Find  a  fourth  proportional  to  the  sides  of  these  squares 
and  AB.  (XVII.  Prob.  I.) 

On  this  fourth   proportional  construct   a  polygon   similar 

to  P.  (XVII.  Prob.  XII.) 

It  will  be  the  polygon  required. 

Q.E.P. 
Proof  is  left  to  the  pupil. 


XVII.     LINEAR   APPLICATION   OF   PKOPORTION        207 


Ex.57.    A  ^CB  is  inscribed.      CT  bisects  ZACB 
pendicular  to  AB. 

Why  does  CT  pass  througli  F  ? 

Wliy  is  FJa.  diameter  of  tlie  circumcircle  ? 

Why  is  F  the  mid-point  of  arc  AB  ? 

Why  is  TF  a  fourtli  proportional  to  JF,  LFy 
and  CF? 

Ex.  58.    Let  x=  TF,   t=zCT,  s  =  LF,   and 
d  =  JF. 

Then  x  td:  is  :x  +  t,  or  x^  +  tx  =  ds.     Solve 

Construct  x. 


FJ  is  a  mid-per- 
GJ 


^(^  ±  V4  L_i  of  d  •  s  +  D  on  0- 


Ex.  59.    By  the  aid  of  the  foregoing,  construct  a  triangle,  having  given 
the  base,  the  vertex  angle,  and  the  bisector  of  the  vertex  angle. 

n 

Ex.  60.    Given,  A  ABC  inscribed  in  a  circle. 
CE  and  CF  are  so  drawn  that 

^ACE=AFCB. 
Prove  that  A  ACE -^  A  FCB. 
Hence,  show  that 

CA'CB=CE'  CF. 

Ex.  61.   In  the  figure,  CF  bisects  ZACB.     Circumscribe  a  circle  to 
A  ABC,  and  produce  Ci^to  L. 

Show  that  AACF-^  LCB. 

Why,  then,  is  AC .  CF.  :  CL:  CB? 

Ex.  62.    Why  does  the  rectangle  of  CF-  FL  -  rectangle  oi  AF-  FB  ? 
Prove,  then,  that  the  rectangle  of  ^C-  CB=  Ci^^  +  rectangleof  J.i''-  FB. 

Ex.  63.  State  the  above  equation  in  general 
terms. 

Ex.  64.  If  LJ  is  a  diameter  of  the  circumcircle 
of  A  ABC,  K  the  center  of  the  incircle,  and  if  KB 
is  drawn,  and  KE±CB,  show  that  Z  ABL  = 
ZLCB. 

Show  that  ZKBA  =  ZKBC,  and  therefore  that 
ZLBK=  ZLKB,  and  therefore  LK=  BL. 

Ex.  65.  Why  is  A  LJB  a  right  triangle  ?  Why 
is  rt.  A  LJB  ~  KEC  ? 

Why,  then,  does  the  □  of  ZJ"  •  KE  =  [:3ot  BL  -  KC  =  n3oi  LK  -  KC? 


208 


THE  ELEMENTS  OF  GEOMETRY 


Ex.  66.  Prove  from  Exercise  66  that  the  rectangle  of  the  diameter  of 
the  circunieircle  of  a  triangle,  and  the  radius  of  the  incircle,  equals  the  rec- 
tangle of  the  segments  of  any  chord  of  the  circumcircle  that  passes  through 
the  center  of  liie  incircle. 

Ex.  67.  If  ^f  is  the  mid-point  of  /TA'i,  and  RA  is  a  perpendicular  to 
A'A'i  at  any  point  R,  then  LK^^  -  UP  =  2  KK\  •  RM.     ( Why  1) 

Prove  that  if  L  be  taken  anywhere  in  tlie 
±RA,  say  at  J, 

JTi^  -  7a^  =  2  KKi .  RM. 
Ex.  68.  Of  what  point,  then,  is  the  ±  RA 
the  locus  ? 

Ex.  69.  In  the  expression 

lA"i^  -  lK-  =  2  im  AT/Ti  •  RM, 
construct  RM. 

(Factor  the  first  member.    XVII.  Prob.  1.) 

Ex.  70.  Find,  then,  the  locus  of  a  point  the  difference  of  the  squares 
of  whose  distances  from  two  given  points  equals  a  given  square. 

Ex.  71.  If  two  circles  are  drawn  with  any  radii,  and  K  and  Ki  as 
centers,  and  trom  any  point  L  in  a  perpendicular  to  KKi,  as  AR,  tan- 
gents are  drawn  to  these  circles,  show  that 

LG^-  LT^  =  2  AVi'i .  RM-(r'^  -  r^). 

Ex.  72.  In  the  second  member  of  the  last  equation,  is  there  any  term 
whose  value  changes  as  L  moves  up  or  down  the  ±AR? 

If,  then,  the  value  of  L^  —  LT^  is  the  same,  or  constant,  no  matter 
where,  in  the  ±AR,  L  may  lie,  construct  the  locus  of  a  point  the  differ- 
ence of  the  squares  of  the  tangents  from  which  to  two  given  circles  equals 
a  given  square. 

Def.    The  Radical  Axis  of  two  circles  is  the  locus  of  a  point  from 
which  the  tangents   to  the  two  circles  are 
equal.  A 

In  such  case  RA  is  called  the  radical  axis  j^ 

of  the  two  circles. 

Ex.  73.^  If  OTT  intersect  QK^  show  that 
the  radical  axis  of  the  two  circles  is  their     ■ 
common  chord  produced. 

Note, — The  radical  axis  of  two  circles  is 
the  locus  of  the  centers  of  circles  which  cut 
the  two  circles  orthogonally. 


XVII.    LINEAR   APPLICATION   OF   PROPORTION 


209 


Ex.74.  In  a  system  of  three  circles  the  radical  axes  concur.  For,  if  tan- 
gents be  drawn  to  the  three  circles  from  the  point  where  two  of  the  radical 
axes  intersect,  these  tangents  will  be  equal.  Therefore,  the  third  radical 
axis  must  pass  through  this  point. 

Def.  This  point  of  concurrence  is  called  the  Radical  Center  of  the  three 
circles.  - 

Thus,  if  RA  is  so  taken  that  LG^  -  LT^  =  0,  then  LG'^  =  LT^,  or 
LG  =  LT.  ....... 


Ex.  75.   If  ^B  is  divided  internally  in  theratio  of  a  :  b  at  the  point  M, 
and  also  divided  externally  at  L  in  the  same 
ratio,  and  if  on  ML  as  a  diameter  a  circle  is 
described,  and  the  point  F  is  any  point  in  the 
circle,  prove  the  following  relations : 

(1)  3IFL  is  a  right  angle. 

(2)  The  rectangle  otAM-  LB  =  the  rectangle 
of  AL .  MB. 

(3)  Prove  that  the  circle  on  ML  as  a  diameter  is  the  locus  of  a  point 
the  ratio  of  whose  distances  from  A  and  B  is  that  of  a  :  Z). 

(4)  If  AM  =  MB,  that  is,  if  the  ratio  a :  6  =  1,  what  becomes  of  the 
point  L  ? 

(5)  If  the  ratio  a  :  6  is  less  than  one,  where  does  the  point  L  reappear  ? 

Ex.  76.  The  apparent  size  of  an  object  is 
determined  by  the  angle  that  it  subtends  at  the 
eye  of  the  observer. 

Thus,  if  at  any  point  Q,  the  tangents  to  two 
unequal  ©  K  and  L  make  equal  angles,  i.e.  if 
/.BQS=  ZHQM,  the  circles  will  appear  equal, 
as  seen  from  Q. 

Draw  the  radii  LB  and  KM,  and  show  that, 
if  Q  be  such  a  poin  t,  A  LQB  is  similar  to  A  KQM. 

Hence,  show  that  if  a  and  b  be  the  radii  of 
the  circles,  QL  :  QK::a:b. 

Hence,  find  the  locus  of  the  point  from  which  two  unequal  circles  seem 
to  be  equal.     (  Ex,  68. ) 

Def.    This  locus  is  called  the  Circle  of  Similitude  of  tlie  given  circles. 


XVIII.     GROUP   ON   CIRCUMSCRIBED   AND 
INSCRIBED   REGULAR   POLYGONS 

DEFINITIONS 

A  Regular  Polygon  is  a  polygon  that  is  both  equiangular  and 
equilateral. 

The  Apothem  of  a  regular  polygon  is  the  radius  of  its  inscribed 
circle. 

The  Radius  of  a  regular  polygon  is  the  radius  of  the  circum- 
scribed circle. 

'The  Center  of  a  regular  polygon  is  the  common  center  of  the 
inscribed  and  circumscribed  circles. 

PROPOSITIONS 

XVIII.  1.  If  a  polygon  is  regular, 

(1)  A  circle  may  he  circumscribed  about  the  polygon. 

(2)  A  concentric  circle  may  be  inscribed  in  the  polygon. 

B.YP-    If  a  polygon 
ABC-E 

is  regular, 

Cone. :  then  (1)  a  circle  may  be  passed  through  A,  B,  O^  etc., 
(2)  a  concentric  circle  may  be  described  tangent 
to  AB,  BO,  etc. 

Dem.  (1)  Pass  a  circle  through  A,  B,  and  C;  let  its  center 
be  K.  Join  K  to  the  vertices  and  to  the  mid-points  of  the 
sides  of  the  polygon  (by  KA,  KB,  etc. ;  KH,  KL,  etc.). 

Kii±Ba 

210 


XV  III.     CIRCUMSCRIBED   AND   INSCRIBED   POLYGONS     211 

[A  radius  ±  to  a  chord  bisects  the  chord,  etc.]  (TX.  1.) 

Kevolve  KHCE  on  KH  as  an  axis,  until  it  falls  on  KHBA. 
The  angles  at  H  are  right  angles,  as  just  shown. 

.-.  HC  takes  the  direction  of  HB.  (Ax.  7.) 

HC=^  HB  (Const.).     .-.  C  falls  on  B. 

A  HCE  =  Z  HBxi  and  CE  =  BA. 

(Def.  of  regular  polygon.) 

.-.  first,  CE  takes  the  direction  BA,  and  second,  E  falls  on  A ; 
i.e,  KA^KE) 

and  the  circle  through  A,  B,  and  C  passes  through  E. 

Similarly,  this  circle  may  be  shown  to  pass  through  any  other 
vertex  of  the  polygon. 

(2)  The  sides  AB,  BC,  etc.,  are  equal  chords  of  the  same  circle. 

.*.  these  sides  are  equidistant  from  the  center. 

[In  the  same  O  equal  chords  are  equidistant,  etc.]      (TX.  2.) 

.-.  A  circle  with  K  as  center  and  a  radius  =  KL  =  KH,  etc., 

is  tangent  to  every  side  of  the  polygon. 

Q.E.D. 

Ex,  1.  In  a  regular  n-gon,  the  central  angle  is  the  supplement  of  any 
one  of  the  interior  angles. 

Ex.  2.  Divide  a  regular  dodecagon  into  triangles  by  drawing  radii, 
Join  any  two  alternate  vertices. 

Prove,  by  finding  the  area  of  the  triangles  crossed  by  the  join,  that : 

The  area  of  a  regular  dodecagon  equals  three  times  the  square  on  the 
radius. 

Ex.  3.  Draw  a  figure  showing  a  circumscribed  equilateral  polygon 
that  is  not  regular. 

Ex.  4.  If  a  circumscribed  polygon  is  equilateral,  the  polygon  will  be 
regular,  provided  the  number  of  sides  be  odd. 

(Use  IX.  5  and  V.  1.) 

Ex.  5.  Explain  why  the  polygon  of  Ex.  4  may  not  be  regular  if  the 
number  of  sides  be  even.  That  is,  show  that  while  there  will  be  two  sets 
of  equal  angles  when  the  number  of  sides  is  even,  the  angles  of  one  set 
will  not  necessarily  be  equal  to  those  of  the  other. 


312         THE  ELEMENTS  OF  GEOMETRY 

XVIII.  2.  Conversely.  If  a  polygon  is  inscribed  in 
a  circle  and  circumscribed  to  a  concentric  circle,  the 
polygon  is  regular,  2? 

Hjrp.  If  a  poly- 
gon ABC-0  is  in- 
scribed in  a  Oe  and 
also  circumscribed  to 
a  concentric  O  i. 

Cone. :  then  the  polygon  is  regular. 

Dem.  AB,  EC,  etc.,  tangents  to  the  O  i,  are  perpendicular  to 
the  radii  KL,  KH,  etc.,  of  the  Or. 

[A  tangent  is  A.  to  the  radius  through  the  point,  etc.]    (IX.  4.) 

But  these  tangents  to  the  O  i  are  chords  of  the  O  e.       (Hyp.) 

Again  the  distances  of  these  chords  from  the  common  center 
of  the  ©  i  and  e  are  equal.   .•.  the  chords  themselves  are  equal. 

[Equal  chords  are  equidistant  from  center,  etc.]         (IX.  2.) 

.*.  the  polygon  is  equilateral. 

Again  the  arcs  AB,  BC,  etc.,  are  all  equal. 

[In  the  same  O,  or  in  equal  (D,  equal  chords,  etc.]       (TX.  3  a.) 

Each  angle  of  the  polygon  intercepts  {n  —  2)  of  these  equal 
arcs,  if  n  be  the  number  of  sides  in  the  polygon. 

.*.  each  angle  of  the  polygon  is  measured  by  ^{n  —  2)  of  these 
equal  arcs. 

[An  inscribed  Z  is  measured  by  one  half,  etc.]         (XII.  3.) 

.*.  each  angle  has  the  same  measure  as  any  other,  n  being 
the  same  for  all. 

.*.  the  polygon  is  equiangular. 

•.•  the  polygon  is  both  equilateral  and  equiangular,  it  is 
regular.  ^^^ 

°  Q.E.D. 

XVIII.  2  a.  The  area  of  a  regular  polygon  equals  one 
half  of  the  product  of  its  perimeter  and  apothem, 

(See  XIII.  3.) 


XVIII.     CIRCUMSCRIBED  AND  INSCRIBED  POLYGONS    213 


XVIII.  3.  If  an  inscribed  polygon  is  equilateral^  the 
polygon  is  regular. 

Prove  as  above  that  the  polygon  is  equiangular. 

XVIII.  L  If  a  regular  hexagon  is  inscribed  in  a  circle, 
the  side  of  the  polygon  equals  the  radius  of  the  circle, 

A 


Hyp.  If  a 
hexagon  AB-G 
is  regular  and 
inscribed  in  a 
OK, 


Cone. :  then  the  side  of  the  hexagon  equals  the  radius  of  the 
circle. 

Dem.  Join  K  to  the  vertices  of  the  hexagon,  and  draw  the 
apothem,  KQ. 

The  A  GKF  is  isosceles.       (KG  =  KF=  radius  of  given  O.) 

Z  GKF=  I  rt.  Z.     (Central  Z  of  a  hexagon.) 

.:.  each  of  the  two  other  angles  =  f  rt.  Z,  and  GKF  is  an 

equilateral  triangle. 

.-.  G^i^=  G^=  the  radius. 

Q.E.D. 

Ex.  6.  Draw  figures  showing  inscribed  equiangular  polygons  that  are 
not  regular. 

Ex.  7.  If  an  inscribed  polygon  is  equiangular,  the  polygon  is  regular, 
provided  the  number  of  sides  be  odd. 

(Use  IX.  3  a  and  XII.  3.) 

Ex.  8.  Explain  why  the  polygon  may  not  be  regular  if  the  number  of 
sides  be  even. 

Ex.  9.    Construct  an  angle  of  4°  30'. 

Ex.  10.    Construct  an  angle  of  72°. 

Ex.  11.    Construct  an  angle  of  24°. 


214 


THE  ELEMENTS  OF  GEOMETRY 


XVIII.  5.  The  radius  is  the  limit  to  which  the  apo- 
them  of  the  inscribed  reguJwr  polygon  approaches,  as 
the  number  of  sides  is  increased  indefinitely. 

Hyp.  If  AB  be  a  side 
of  a  regular  w-gon  of 
apothem.  KLy 

Cone. :  then  KL  =  KA  as  n  is  indefinitely  increased. 

Dem.  AK-  KL  <  AL.  (VII.  2  a. ) 

AL=^\AB. 

As  n  increases,  AB  diminishes,  and  may  be  made  as  small  as 

^®  P^^^s®-  .-..  AL  =  0.  (Def.  of  limit.) 

.*.  AK—  KL  =  0 ;  (Same  reason.) 

i.e.  KL  =  AK.  (Same  reason.) 

Q.E.D. 

XVIII.  5  a.  The  radius  of  a  regular  circumscribed 
n-gon  approaches  as  a  limit  the  radius  of  the  inscribed 
circle  as  n  is  indefinitely  increased.  x — ^r.i  ^^ — yM 

Let  the  student  supply  the  proof,  which  is 
exactly  similar  to  the  above,  using  the  adjoin- 
ing figure.  JC 

XVIII.  5  b.  The  square  on  the  apothem  approaches 
as  a  limit  the  square  on  the  radius.       (Fig.  of  Theorem.) 

XVIII.  6.  The  circumference  is  the  common  limit  to 
which  the  perimeters  of  similar  inscribed  and  circum- 
scribed regular  polygons  approach  as  the  number  of 
sides  is  increased  indefinitely. 

Hyp.  If  C  is  the  circumference  of  a  circle,  of  radius  r,  and 
P  and  F  are  the.  perimeters  of  the  regular  circumscribed  and 
similar  inscribed  n-gons, 


XVIII.     CIRCUMSCRIBED  AND   INSCRIBED  POLYGONS     216 

Cone. :  then,  as  n  is  indefinitely  increased, 

P=C  and  P'  =  (7. 

Dem.  Let  AB  and  EH  be  the  sides  of  the  n-gons ;  KM  and 
KL,  the  apothems. 

Then  P.P.:  KM :  KL.  (XVII.  2.) 

.-.  P-P'.P.'.  KM-  KL  :  KM.  (XI.  1.  Sch.  Cone.  (4).) 

.-.  (P- P')  '  KM=  P'  (KM-  KL)  =  P •  {r- KL).   (XL  1.) 

.-.  P-P'=P'  ^'~^^^.  (Ax.  3.) 

/Of  ^  ^ 

But  r  -  KL  =  0  (XVIII.  5.), 

and.  P  decreases  as  n  increases.  (Why  ?) 

...  p.rzi^  =  0;    i.e.  P-P'  =  0.  (Def.  Limit.) 

.  Now  C>P',  and  (7<P.  .      (Why?) 

.-.  P-C<P-P'. 

.-.  since       P-P'  =  0,  P-C=0,  and  P=  C. 

Similarly,  C-P'<P-P\  C-P'  =  0  and  P=a 

Q.E.D 

XVIII.  6  a.  7%e  czVcZe  is  the  common  limit  to  which, 
the  areas  of  similar  inscribed  and  circumscribed  regular 
polygons  approach  as  the  number  of  sides  is  increased 
indefinitely. 

XVIII.  7.  The  area  of  a  circle  equals  one  half  the 
product  of  its  circumferejice  and  radius. 

Let  the  pupil  supply  the  proof  by  using  2  a  and  6  a. 

Ex.  12.    Construct  an  angle  of  6°. 

Ex.  13.  The  perimeter  of  an  inscribed  square  is  40  ft.  What  is  the 
radius  of  the  circle  ? 

Ex.  14.  The  perimeter  of  a  square  is  40  ft.  What  is  the  length  ot 
the  apothem  of  the  square  ? 

What  is  the  length  of  the  radius  of  the  square  ? 


216  THE   ELEMENTS  OF   GEOMETRY 

XVIII.  SUMMARY  OP  PROPOSITIONS  IN  THE  GROUP 
ON  CIRCUMSCRIBED  AND  INSCRIBED  REGULAR 
POLYGONS 

1.  If  a  polygon  is  regulaVy 

(1)  A  circle  maj/  he  circumscribed  about  the  polygon; 

(2)  A  concentric  circle  may  be  inscribed   in   the 

p)olygon, 

2.  Conversely.  If  a  polygon  is  inscribed  in  a  circle 
a7id  circumscribed  to  a  concentric  circle,  the  polygon  is 
regular. 

a  The  area  of  a  regidar  polygon  equals  one  half 
of  the  product  of  its  perimeter  and  apothem. 

3.  If  an  inscribed  polygon  is  equilateral,  the  polygon 
is  regular. 

4:.  If  a  regular  hexagon  is  inscribed  in  a  circle,  the 
side  of  the  polygon  equals  the  radius  of  the  circle. 

5.  The  radius  is  the  limit  to  ivhich  the  apothem  of  the 
inscribed  regidar  polygon  approaches  as  the  number  of 
sides  is  increased  indefinitely. 

a  The  radius  of  a  regular  circumscribed  n-gon 
approaches  as  a  limit  the  radius  of  the 
inscribed  circle  as  n  is  indefinitely  increased. 

b  The  square  on  the  apothem  approaches  as  a. 
limit  the  square  on  the  radius. 


XVIII.     CIRCUMSCRIBED   AND  INSCRIBED   POLYGONS     217 

6.  The  circumference  is  the  common  limit  to  which 
the  perimeters  of  similar  inscribed  and  ciixumscrihed 
regular  polygons  approach  as  the  number  of  sides  is 
increased  indefinitely. 

a  The  circle  is  the  common  limit  to  which  the 
areas  of  similar  inscribed  and  circumscribed 
regular  polygons  approach  as  the  number  of 
sides  is  increased  indefinitely. 

7.  The  area  of  a  circle  equals  half  the  product  of  its 
circumference  and  radius. 

PROBLEMS 

Prob.  I.  To  inscribe  a  regular  hexagon  in  a  given 
circle. 

Given.     TheO^. 

Required.     To  inscribe  a  regular  hexagon 
in  O  K. 

Analysis.     Suppose  AB  is  a  side   of  a 
regular  hexagon  inscribed  in  OK. 

Then  AB  =  AK,  the  radius  of  the  circle.  (XVIII.  4.) 

Hence,  the  construction. 

Const.  With  any  point  in  the  circumference  as  a  center,  ap- 
ply the  radius  of  the  circle  from  this  starting  point  six  times 

as  a  chord ;  a  regular  hexagon  is  thus  formed. 

Q.E.D. 

Note  1. — The  joins  of  the  alternate  points  will  form  an  inscribed 
equilateral  triangle. 

Note  2.  — The  joins  of  the  vertices  of  the  regular  hexagon,  with  the 
mid-points  of  the  arcs  subtended  by  the  sides  of  the  hexagon,  will  form  a 
regular  inscribed  dodecagon. 


218  THE   ELEMENTS  OF  GEOMETRY 

Prob.  II.  To  inscribe  a  regular  decagon  in  a  given 
circle. 

Given.     The  O  K. 

Required.     To  inscribe  a  regular  decagon 
in  O  K 

Analysis.     Suppose  a;  is  a  side  of  a  regu- 
lar inscribed  decagon. 

If  a;  be  a  side  of  a  regular  inscribed 
decagon,  then  if  radii  KA  and  KB  be 
drawn,  /.  K=  -^  oi  A  rt.  A,  or  |  rt.  Z,  and  each  base  angle  = 
I  rt.  Z.  Again,  if  BG  bisect  Z  ABK,  then  A  ABC  and  A  BCK 
are  isosceles. 

Now  A  ACB  is  similar  to  A  AKB.  (XV.  4.) 

.*.  KB  (or  r)  :  AB  (or  x)  : :  BC  (or  its  equal,  x)  :  AG. 

(Def.  of  ~  w-gons.) 

That  is,  r:x::x:  AG, 

or  X  is  the  greater  segment  of  the  radius  divided  in  extreme 
and  mean  ratio. 

Suggested  theorems,  etc.,  are  IV.  Ex.  18,  and  XV.  2. 

Applicable  theorems :  both  the  above. 

Const.     Divide  the  radius  KA  in  extreme  and  mean  ratio. 

The  greater  segment  is  the  side  of  the  require'd  regular 
decagon. 

Proof.         r:x::x:CA,  or  KA  :AB::  AB  i  GA.         (Const.) 

.-.  KAB  is  similar  to  AGB.  (XV.  4.) 

.*.  AGB  is  isosceles,  and  BGK  is  isosceles. 

.-.  Z  K=  Z  GBK=  i  AGB ;  (III.  2  a.) 

that  is,  ZK=IKAB. 

.:  ZK=^Tt.Z.     ' 

^  Q.E.D. 

Ex.  15.  Find  the  cost,  at  $  2..30  a  yard,  of  building  a  stone  wall  around 
a  lot,  in  the  shape  of  a  regular  hexagon,  containing  260  sq.  yds. 

Ex.  16.  The  perimeters  of  a  regular  hexagon  and  a  regular  octagon  are 
each  240  ft.    What  is  the  difference  in  their  areas  ? 


XVIII.     CIRCUMSCRIBED   AND   INSCRIBED  POLYGONS     219 

Ex.  17.  What  is  the  area  of  a  garden  in  the  shape  of  a  regular  deca- 
gon, one  side  of  which  is  18  ft.  long  ? 

Ex.  18.   The  perimeter  of  a  regular  hexagon  is  42  ft.     What  is  its  area  ? 

Ex.  19.  The  perimeter  of  a  regular  hexagon  is  30  ft.  What  is  the 
length  of  the  radius  of  the  hexagon  ? 

Ex.  20.  Two  regular  octagons  contain  108  sq.  ft.  and  96  sq.  ft.,  respec- 
tively. What  is  the  length  of  the  side  of  a  third  regular  octagon  equal  in 
area  to  the  sum  of  the  first  two  ? 

Ex.  21.  A  regular  decagon  is  inscribed  in  a  circle  whose  radius  is  10  ft. 
Find  its  area.     (Use  Prob.  II,  p.  218.) 

Ex.  22.  A  lawn  in  the  shape  of  a  regular  octagon  measures  186  ft. 
on  each  side.     What  is  its  area  ? 

Ex.  23.  If  the  sides  of  three  regular  decagons  are  3  ft.,  4  ft.,  and  12  ft., 
respectively,  what  is  the  side  of  a  regular  octagon  whose  area  is  equal  to 
the  sum  of  the  three  given  regular  decagons  ? 

Let  r  be  the  radius  of  any  regular  polygon,  s  the  side  of  the  polygon,  p 
the  apothem,  and  A  the  area.     Show  that : 

Ex.  24.    In  a  square  p  =  -  V2  ;  J.  =  2  r^. 

Ex.  25.   In  an  equilateral  triangle p  =  -;  A  =  '—^ VS. 

r    /-  3  r^    /— 

Ex.  26.    In  a  regular  hexagon  p  =  -  v3  ;  ^  =  -^  V3. 

2  2  _ 

Ex.  27.    In  an  equilateral  triangle  r  =  2p\  s  =  2pVs  ;  A  =  I  s^VS. 

Ex.  28.    In  a  regular  octagon  s  =  r V 2  —  V2. 

Ex.  29.    In  a  regular  decagon  s  =  -  (  V5  —  1). 

Ex.  30.   In  a  regular  hexagon  A  =  -^  V3. 

Ex.  31.  In  a  regular  dodecagon  ^  =  3  r^. 

Ex.  32.    In  a  regular  octagon  A  =  2  r^-\/2. 

Ex.  33.  The  square  on  the  side  of  the  inscribed  equilateral  triangle 
equals  three  times  the  square  on  the  side  of  the  inscribed  regular  hexagon. 

Note.  —Let  V2  =  1.414,  V3  =  1.732,   V5  =  2.236. 

Ex.  34.    In  a  square  r  =  14  ft.     Find  p  and  A. 

Ex.  35.  Find  the  area  of  an  equilateral  triangle  inscribed  in  a  circle 
whose  radius  is  10  ft. 

Ex.  36.    Find  the  apothem  of  a  regular  hexagon  whose  side  is  12  meters. 

Ex.  37.    Find  the  side  of  an  equilateral  triangle  whose  apothem  is  6  ft. 

Ex.  38.    Find  the  radius  of  a  regular  octagon  whose  side  is  4  ft. 

Ex.  39.    What  is  the  side  of  a  regular  decagon  whose  radius  is  10  ft.  ? 

Ex.  40.  The  area  of  a  regular  hexagon  is  1732  sq.  ft.  What  is  the 
length  of  each  side  ? 


XIX.  GROUP  ON  THE  AREA  OF  THE  CIRCLE 

PROPOSITIONS 

XIX.  1.   The  circumferences  of  two  circles  are  to  each 
other  as  their  radii  and  as  their  diameters. 

Hyp.  If  the  circum- 
ferences of  any  two 
circles  be  denoted  by 
C  and  C",  and  their 
radii  by  r  and  r', 

Cone:  then  ^=:^'  and    ;^  =  -^. 

r      r'  2r     2r' 

Dem.     Inscribe  in  the  circles  regular  polygons  of  the  same 
number  of  sides. 

These  two  polygons  will  be  similar.  (XV.  8.) 

Denoting  the  perimeters  of  the  polygons  by  P  and  P'  we  have 

^  =  ^.  (XVII.  2.) 

r      r 

Now,  if  the  number  of  sides  be  increased  indefinitely,  P  and 

P  P' 

P  will  approach  their  limits  (7  and  C".     That  is,  —  and  — -, 

r  r' 

C         O 
while  remaining  equal,  will  approach  as  their  limits  —  and  —  • 

...  ^=^.  (Postulate  of  Limits.) 

r      r 

Dividing  by  2  we  have 


2  r     2/ 
220 


Q.E.D. 


XIX.     AREA   OF  THE   CIRCLE  221 

XIX.  1  a.  The  ratio  of  the  circumference  to  the  diam- 
eter is  constant. 

Dem.  If  C  be  the  first  circumference  and  C  the  second, 
and  r  and  r'  the  respective  radii, 

then  ^  =  Fi'  (XIX.  1.) 

2r     2r'  ^  ^ 

C 
That  is,  the  ratio  —  is  the  same  whatever  the  size  of  the 
2r 

circle.     In  other  words, 

C  -^  2  9'  is  a  constant. 

Q.E.D 

ScH.     This  constant  is  called  tt. 

[tt  is  the  initial  letter  of  the  Greek  word  for  perimeter.] 
Hence,  -  —  =  tt,  or  C  ==  2  irv. 

If  r  =  l,       ^=|. 

XIX.  2.   The  area  of  a  circle  is  -n-r^. 

Hyp.     If  K 
be     a     circle  (        /c 

of  radius  r, 

Cone. :  then  the  area  of  this  circle  =  tt?--. 
Dem.     The  area  of  the  circle  =  O  x  |-  r. 
[The    circumference   is   the    common    limit   to   which    the 
perimeters,  etc.]  (XVIII.  6.) 

But  the  circumference  =  2  ttt.  (XIX.  1.  Sch.) 

.  •.  the  area  of  the  circle  =  2  Trr  x  ^ 

—  T^^'  «  „  ^ 

Q.E.D. 

XIX.  2  a.  The  areas  of  two  circles  are  to  each  other 
as  the  squares  of  the  radii  and  as  the  squares  of  the 
diameters. 


'700 


THE  ELEMENTS  OF  GEOMETRY 


XIX.  3.  Prob.  Given  the  length  of  a  side  of  a  regular 
polygon,  inscribed  in  a  circle  of  radius  r,  to  find  the 
side  of  a  regular  polygon  of  double  the  number  of  sides, 
inscribed  in  the  same  cbxle.  a 

B 


Given.     The  value  a  of  a  side  AB  of  a    ^ 
regular  ?i-goii   inscribed   in   a   circle   of   ra- 
dius r. 

Required.     To  find  the  side  of  the  regular 
(2  n)-gon  inscribed  in  the  same  circle. 

Const.     Draw  FKE  the  mid-perpendicular  of  AB^  intersect- 
ing the  circumference  at  (7;  draw  AG.     Draw  ^/fand  AF. 
AC  will  be  the  side  of  the  regular  (2  n)-gon.  (Why  ?) 

It  is  required  to  compute  the  value  of  AC  in  terms  of  a. 

Computation.    In  the  vt.AAEK,  EK^=AK'-AE^(XlY,la.) 


that  is, 


Again, 


But 


EK'  =  ?-2  _ 


4^ 


EK 


=  V^^^  =  iV4l^-3^ 


Z  CAF  is  a  right  angle. 
AC'^CF-  CE. 


(XII.  2  a.  Sch.) 
(XVII.  5  6.) 


CF=2r;  GE  =  GK  -  EK=r  -  EK. 
AC''  =  GF'CE  =  2  r(r  -  EK) 

2r( 

2  7-2 


=  2r(r-iV4r2-a2) 
r  V4  7^  —  a^. 


Q.E.F. 
ScH.     The   perimeter  of  the  regular   2  n-gon  =  2  nAG  ^ 
2  n  V  2  r^  —  r V4 1^  —  d\     It  is  most  convenient  to  take  r  =  1, 
when  the  expression  becomes  2  n^2  —  V4  —  aK 


XIX.     AREA  OF  THE   CIRCLE  223 

XIX.  4.  Prob.    To  compute  the  value  of  tt  approxi- 
mately. 
Sol.     ,r  =  —  (XIX.  1.  Sch.).    Hence,  if  r  =  1, 

Let  P„  denote  the  perimeter  of  a  polygon  of 
n-sides. 

Beginning  with  the  value  n  =  6,  we  find  by  XIX.  3.  Sch. : 

Pe    =6. 

Pi2  =  6.2116571 

P24  =  6.2652572 

P48  =  6.2787004 

P96  =  6.2820640 

Pi92  =  6.2829051 

P384  =  6.2831154 

P768  =  6.2831694 

We  may  continue  this  operation  as  far  as  we  please,  but  for 

all  practical  purposes  the  perimeter  of  the  polygon  of  768  sides 

coincides  with  the  circle. 

.-.  7r  =  6.283169-^2  =  3.14159  (nearly). 

Q.E.F. 

Note. — Lambert  (1750)  proved  that  tt  is  incommensurable  with  1. 
Lindeniann  (1882)  went  further  and  showed  that  tt  cannot  be  expressed 
algebraically. 

Ex.  1.  The  minute  hand  of  a  tower  clock  is  6  ft.  long.  What  is  the 
length  of  the  circumference  of  the  clock  ? 

Ex.  2.    What  is  the  area  of  the  face  of  the  clock  ? 

Ex.  3.    What  is  the  circumference  of  a  circle  whose  diameter  is  10  in.  ? 

Ex.  4.  What  is  the  diameter  of  a  circle  whose  circumference  is  27  ft. 
8  in.  ? 

Ex.  6.  The  minute  hands  of  two  tower  clocks  are  respectively  6  ft.  and 
8  ft.  long.     What  is  the  ratio  between  the  lengths  of  the  circumferences  ? 

Ex.  6.    What  is  the  ratio  between  their  areas  ? 

Ex.  7.  What  is  the  length  of  an  arc  of  36°  in  a  circle  whose  diameter 
is  24  in.  ? 

Ex.  8.    What  is  the  diameter  of  a  circle  whose  area  is  40  A.  ? 


224  THE  ELEMENTS  OF  GEOMETRY 


XIX.    SUMMARY   OF  PROPOSITIONS  IN  THE  GROUP 
ON   THE  AREA   OF   THE   CIRCLE 

1.  The  circumferences  of  two  circles  are  to  each  other 
as  their  radii  and  as  their  diameters. 

a    The  ratio  of  a  circumference  to  its  diameter  is 
constant. 
ScH.    The  circumference  of  a  circle  is  2  rrr. 

b  The  areas  of  any  two  circles  are  to  each  other 
as  the  squares  of  the  radii,  and  as  the  squares 
of  the  diameters. 

2.  The  area  of  a  circle  is  wr*. 

3.  Prob.  Given  the  length  of  a  side  of  a  regular 
polygon,  inscribed  in  a  circle  of  radius  r,  to  find  the 
side  of  a  regular  polygon  of  double  the  number  of  sides, 
inscribed  in  the  same  circle. 

ScH.  Application  of  the  foregoing  problem  to  the  calcula- 
tion of  perimeters. 

4  Prob.    To  compute  the  value  of  v  approximately. 


XIX.     AREA  OF  THE   CIRCLE  225 

Ex.  9.  The  radius  of  a  circle  is  8  ft.  What  is  the  radius  of  a  circle 
100  times  as  large  ? 

Ex.  10.  A  wheel  makes  420  revolutions  in  traveling  half  a  mile.  What 
is  its  diameter  ? 

Ex.  11.  To  find  a  circle  whose  circumference  is  two  thirds  a  given 
circumference. 

Ex.  12.  A  ten-inch  water  pipe  discharges  200  gallons  a  minute.  What 
is  the  diameter  of  a  pipe  that  discharges  800  gallons  a  minute  under  the 
same  pressure  ? 

Ex.  13,  A  circular  pipe  10  in.  in  diameter  delivers  100  gallons  per 
minute.  If  the  capacity  is  to  be  increased  fourfold,  what  will  be  the 
diameter  of  the  new  pipe  ? 

Ex.  14.  The  radius  of  a  circle  is  12  in.,  and  the  length  of  an  arc  is  the 
same.     What  is  the  angle  subtended  at  the  center  by  the  arc  ? 

Ex.  15.  The  minute  hand  of  a  clock  is  6  ft.  long.  How  long  is  the  arc 
described  by  the  hand  in  10  min.  ? 

Ex.  16.  What  is  the  ratio  between  the  central  angles  of  two  circles  of 
unequal  radii  ? 

Ex.  17.  In  a  clock  whose  minute  hand  is  6  ft.  long,  how  many  degrees 
in  the  central  angle  that  subtends  an  arc  10  ft.  long  ? 

Ex.  18.  Having  found  the  value  of  the  central  angle  in  the  pre- 
ceding question,  by  what  proportion  may  you  determine  the  central 
angle  subtending  an  arc  of  10  ft.  in  a  clock  whose  ininute  hand  is  8  ft. 
long? 

Ex.  19.  In  a  square  closet  whose  side  measures  28  ft.  is  to  be  made  a 
circular  shelf  1  ft.  wide,  with  its  circumference  touching  the  walls  of  the 
room.     Find  the  area  of  the  shelf. 

Ex.  20.  The  area  of  a  circular  mirror  is  314  sq.  in.  The  frame  is  5  in. 
wide.     If  TT  =  2^,  what  is  the  area  of  the  frame  ? 

Ex.  21.  To  construct  a  circle  equal  to  the  difference  between  two  given 
circles  of  radii  a  and  6,  respectively. 

Ex.  22.  To  construct  a  circle  equal  to  the  sum  of  two  given  circles,  of 
radii  a  and  6,  respectively. 


226  THE  ELEMENTS  OF  GEOMETRY 

Ex.  23.  To  construct  a  circle  equal  to  the  sum  of  several  circles  of 
radii  a,  6,  c,  e,  etc. 

Ex.  24.  The  perimeters  of  an  equilateral  triangle,  a  square,  and  a  circle 
are  each  equal  to  204  ft.     Compare  the  areas  of  tlie  three  figures. 

Ex.  25.  The  area  of  a  square  is  266  sq.  ft.  What  is  the  area  of  the 
circle  inscribed  iu  the  square  ? 

Ex.  26.  What  is  the  ratio  of  the  area  of  a  circle  to  that  of  the  in- 
scribed regular  hexagon  ? 

Ex.  27.  What  is  the  area  of  a  circle  if  the  area  of  the  inscribed  regular 
hexagon  is  17.32  sq.  ft.  ? 

Ex.  28.  The  span  (chord)  of  an  arch  in  a  doorway  in  the  form  of  a  cir- 
cular arc  is  26  ft. ,  and  its  height  above  the  stone  piers  is  5  ft.  What  is  the 
radius  of  the  circle  ? 

Ex.  29.  The  altitude  of  a  segment  is  3  ft.  ;  the  radius  of  the  circle  is 
6  ft.  2  in.     Find  the  base  (chord)  of  the  segment. 

Ex.  30.  The  length  of  an  arc  is  42  ft.,  and  the  radius  of  the  circle  is 
29  ft.     What  is  the  area  of  the  sector  ? 

Ex.  31.  In  a  given  equilateral  triangle,  describe  three  circles  tangent 
to  the  sides  of  the  triangle  and  to  each  other. 

Ex.  32.   Semicircles  are  described  on  the  sides 
of  a  right  triangle  as  shown  in  the  figure. 
What  is  the  expression  for  the  area  of  each  ? 

Ex.  33.  How  does  the  sum  of  the  areas  of  the 
smaller  semicircles  compare  with  the  area  of  the 
largest  ? 

Ex.  34.  If  a  segment  AGC  be  subtracted  from  the  semicircle  on  AC^ 
what  area  remains  ? 

If  segment  CEB  be  subtracted  from  the  semicircle  on  J5C,  what  area 
remains  ? 

Ex.  35.    Hence,  show  that  area  rt.  A  A  CB  =  area  AHCG  +  area  BFCE. 


Note.  — These  curvilinear  figures  are  called  the  Lunulas  of  Hippocrates 
(470  B.C.). 

This  is  the  first  equation  established  between  rectilinear  and  curvilinear 


XIX.     AREA   OF  THE   CIRCLE 


227 


Ex.  36.    In  the  figure,  the  diameter  AE  =  6,  AB  =  2,  and  BE  =  4. 

What  is  the  ratio  of  the  circumference  on  AE  to  that  on  BE  ? 

Ex.  37.  What  is  the  ratio  of  the  semicircum- 
ference  on  AE  to  that  on  BE  ? 

Ex.  38.  What  is  the  length  of  the  diameter  of 
a  circumference  equal  to  the  circumference  on  BE 
plus  the  circumference  on  AE  ? 

E-K.  39.  Prove  that  the  semicircumference  on 
AE  equals  the  sum  of  the  semicircumferences  on 
AB  and  BE;  that  is,  that  the  curve  AMBCE 
equals  the  semicircumference  on  AE. 

Ex.  40.    What  is  the  area  of  AMB  ?    Of  BCE  ? 

Ex.  41.    How  much,  then,  is  added  to  the  upper  semicircle  by  AMB  ? 
.How  much  is  subtracted  from  the  same  semicircle  by  BCE  ? 

Ex.  42.   What,  then,  is  the  area  of  the  shaded  horn  ? 

Ex.  43.    What  is  the  area  of  the  unshaded  horn  ? 

Ex.  44.  In  the  figure  the  diameter  is  divided 
into  six  equal  parts. 

Prove  that  the  contour  or  boundary  line  of  any 
one  of  the  figures  equals  that  of  any  other — the 
figures  to  lie  between  two  consecutive  lines. 

Ex.  45.   Similarly  with  the  areas. 


XX.      GROUP   ON   CONCURRENT   TRANSVER- 
SALS  AND   NORMALS 

PROPOSITIONS 

XX.  1.  If  three  transversals  through  the  vertices  of  a 
triangle  are  concurrent,  the  product  of  one  set  of  three 
alternate  segments  determined  hy  the  transversals  on 
the  sides  of  the  triangle  equals  the  product  of  the  other 
set,  and  conversely. 


Hyp.    If,  in  the  A  ABC, 
AE,  BF, 
aud  CG 
concur, 


"^B 


Cone. :  then     GB  •  EC  •  FA  =  AG  -  BE  -  OF. 

Dem.     Draw  AL  and  BM,  the  altitudes  of  A  AOC  and  BOC, 
respectively. 

The  base  CO  is  common  to  the  triangles. 

Then  AAOC:  ABOC: :  AL:  BM.  (1) 

[A  with  equal  bases  are  to  each  other,  etc.]    (XIII.  1  c.  Sch.  1.) 

But  AALGr^ABMG. 

[They  are  right  triangles,  and  Z LGA  =  Z BGM.~\   (XV.  2  a.) 

•228 


XX.     CONCURRENT  TRANSVERSALS  AND  NORMALS     229 

(2) 

(3) 
(4) 


.-.  AL'.BM: 

:AG: 

GB. 

.'. 

AAOC.ABOC: 

.AG: 

GB, 

AAOC 

AG 

or 

A  BOG 

GB 

Similarly, 

A  BOA 
A  GO  A 

BE 

-EG' 

and 

A  GOB 
AAOB 

GF 
'FA 

Multiplying  the  equations  (3),  (4),  and  (5)  together,  member 
by  member,  and  canceling  in  the  first  member,  we  obtain 

H^AG    BE     GF 
GB  '  EG  '  fa' 


Q.E.D. 


or  GB  '  EG  -FA^AG  '  BE  '  GF. 

Conversely.     If 

GB'  EG  '  FA  =  AG  '  BE  '  GF, 
Cone. :  then  AE,  BF,  and  GG  concur. 

Dem.  If  the  transversals  do  not  concur,  at  least  two  of  them, 
say  BF  and  CG,  will  meet,  say  at  0. 

Join  A  and  0,  and  suppose  that  this  join,  instead  of  meeting 
BG  in  E,  meets  it  in  E'. 

.-.  £^  =  ^,  which  is  impossible,  for  if  ^^'>^^,  then 
E  C      EC 
E'C<EC  (see  figure),  and  the  first  fraction  is  greater  than 
the  second.    If  BE'  <  BE,  then  E'C  >  EC,  and  the  first  fraction 
is  less  than  the  second. 

.-.  E'  must  coincide  with  E,  and  AE,  BF,  CG  concur. 

Q.E.D. 


230         THE  ELEMENTS  OF  GEOMETRY 

XX.  2.  Three  concurrent  perpendiculars  divide  the 
sides  of  a  triangle  so  that  the  sum  of  the  squares  of 
one  set  of  alternate  segments  equals  the  sum  of  the 
squares  of  the  other  set,  and  conversely. 

Hyp.  If  /    ^^, 

OE,    OF,    OG 

are  concurrent  F/ 

J§  on  the  sides 

of  the  A  ABC,  

Cone:  then  AS"  +  BE"  +  CF^  =  G^  +  EG^  +  Fj". 
Dem. .  Join  0  with  A,  B,  and  G, 

A0>--AG'=GG'  =  B0'--  GS,     (XIV.  1  a,  and  Ax.  1.) 
Similarlj  on  the  other  sides. 

.-.  ag^-bg'+be^-ec'  +  cf^-Fa'= 

AO'-Bd'-\-BO'-C&+CO'-AO'  =  0. 
Transposing, 

ag'+be''-\-gf'  =  gb^+ec^+f3'. 

Conversely.    If 

AG^  +  BE'-{-CF''=GB'  +  EC^fFA', 

Cone. :  then  perpendiculars  to  the  sides  of  the  triangle  at  E, 
F,  and  G  concur. 

Dem.  Suppose  the  perpendiculars  do  not  concur ;  that  those 
at  F  and  E  meet  at  0,  while  the  perpendicular  from  0  to  AB 
meets  AB  at  G'. 

Then       AG''  +  BE'+CF'=G^  +  EC'  +  FA'.  (1) 

(By  the  direct  Th.) 

AG'  +  BE'-^CF^=GB'-\-EC'-hFA\  (2) 

(By  Hyp.) 


Q.E.D. 


XX.     CONCURRENT  TRANSVERSALS   AND   NORMALS     231 

Subtracting  (2)  from  (1),  member  from  member,  . 

which  is  impossible  unless  G=  G'-,  f or  if  (r  ^  G',  the  first 
member  is  +  and  the  second  is  — ,  or  vice  versa,  and  a  + 
quantity  cannot  equal  a  —  quantity. 

.-.  the  perpendicular  from  0  on  AB  must  fall  at  G;  i.e.  the 
three  perpendiculars  to  the  sides  at  E,  F,  and  G  must  concur. 

Q.E.D. 

Ex.  1.    Show,  by  using  XX.  1,  that  the  following  sets  of  angle-trans- 
versals of  a  triangle  are  concurrent : 
(a)  The  medians. 
(6)  The  altitudes. 

(c)  The  joins  of  the  vertices  to  the  points  of  contact  of  the  inscribed 
circle. 

(d)  The  bisectors  of. the  interior  angles. 

(e)  The  bisectors  of  two  exterior  angles  and  an  interior  angle  not 
adjacent  to  either  of  these  exterior  angles. 

Ex.  2.    Show  that  the  perpendiculars  erected  to  the  sides  of  a  triangle 
at  the  points  of  contact  of  the  escribed  circles  are  concurrent. 

Ex.  3.    Show  from  XX.  2  that  if  each  of  three  circles  intersect  both 
the  others,  the  three  common  chords  are  concurrent. 


282  THE  ELEMENTS  OF  GEOMETRY 


XX.     SUMMARY  OP  PROPOSITIONS  IN  THE  GROUP  ON 
CONCURRENT  TRANSVERSALS  AND  NORMALS 

1.  If  three  transversals  through  the  vertices  of  a  tri- 
angle are  concurrent,  the  product  of  one  set  of  three 
alternate  segments  determined  by  the  transversals  on 
the  sides  of  the  triangle  equals  the  product  of  the  other 
set,  and  conversely. 

2.  Three  concurrent  perpendiculars  divide  the  sides 
of  a  triangle  so  that  the  sum  of  the  squares  of  one  set 
of  alternate  segments  equals  the  sum.  of  the  squares  of 
the  other  set,  and  conversely. 


SOLID    GEOMETRY 

XXI.     GROUP   ON   THE   PLANE   AND   ITS 
RELATED   LINES 

DEFINITIONS 

A  Plane  has  already  been  defined  to  be  a  surface  such  that  if 
any  two  of  its  points  be  joined  by  a  straight  line,  this  line  will 
lie  wholly  within  the  surface. 

A  plane  is  said  to  be  determined  when  it  fulfills  such  condi- 
tions that  its  position  is  fixed. 

No  two  planes  can  fulfill  the  same  set  of  determining  condi- 
tions without  coinciding  throughout  their  whole  extent. 

Corollaries  of  the  Definition 

(a)  A  straight  line  and  a  point  without  the  line  deter- 
mine a  plane. 

Dem.     Let  AB  be  the  given  line  j^ 

and  C  the  given  point.  4^::r::^r2\ 

Let  MQ  be  any  plane  through  AB.        yS^"^^    \      ,^t^^ 

Revolve  MQ  on  AB  as  an  axis,  ^'^^^^^b'        "^(i 

until  it  contains  C. 

Let  this  position  of  the  plane  be  Bh. 

The  plane  BL  is  fixed. 

For  if  it  be  revolved  in  either  direction  about  AB^  it  will  no 

longer  contain  C 

Q.E.D. 


284 


THE  ELEMENTS  OF  GEOMETRY 


(6)  Three  points  nob  in  the  same  straight  line  deter- 
mine  a  plane. 

Dem.     Join  any  two  of  the  points;  apply  Cor.  (a). 

(c)  Two  intersecting  lines  determine  a  plane. 

{d)  Two  parallels  determine  a  plane. 

(e)  The  plane  determined  hy  a  line  and  a  point  is 
identical  with  tJie  plane  determined  hy  this  line  and 
the  parallel  to  it  thai  contains  tJie  given  point. 

(/)  A  straight  line  cannot  intersect  a  plane  in  more 
than  one  point. 

(g)  If  four  points^  A,  B,  C,  and  E,  are  not  in  the  same 
plane  (i.e.  are  not  coplanar),  no  three  of  the  points  can 
he  collinear. 

Dem.     If,  for  example,  AyB,  and  C  lie  in  the  same  straight 

line,  then  this  line  determines  with  E  a  plane,  i.e.  all  four 

points  are  coplanar,  which  contradicts  the  hypothesis. 

.-.  no  three  of  the  points  can  be  collinear. 

^  Q.E.D. 

The  point  in  which  a  line  meets  a  plane  is  called  the  Foot  of 
the  Line. 

A  Perpendicular  to  a  Plane  is  a  line  perpendicular  to  every 
line  of  the  plane  that  passes  through  its  foot. 

The  Projection  of  a  Point  on  a  plane  is  the  foot  of  the  perpen- 
dicular from  the  point  to  the  plane. 

The  Projection  of  a  Line  (straight  or  curved)  on  a  plane  is  the 
locus  of  the  projections  of  the  points  of  the  line. 

The  Angle  that 
a  straight  line 
makes  with  a  plane 
is  the  angle  formed 
by  the  line  and  its 
projection  on  the 
plane. 

A  line  is  parallel  to  a  plane  when  any  plane  through  the 
line  intersects  the  given  plane  in  a  line  parallel  to  the  given 
line. 


,^-^^1^ 


XXI.     THE   PLANE   AND   ITS   RELATED   LINES        235 
PROPOSITIONS 

XXI.  1.    The  line  of  intersection  of  two  planes  is 
straight.    ^^^^^^^^^^^^^^^^^^^^ 


Hyp.     If  plane  CF  intersects  plane  EG  in  the  line  AB, 

Cone. :  then  AB  is  a  straight  line. 

Dem.  If  AB  is  not  a  straight  line,  it  must  contain  at  least  a 
third  point  that  is  not  in  the  same  straight  line  with  A  and  B. 

But  three  points  not  in  the  same  straight  line  determine  the 
position  of  a  plane.  (Def.  of  Plane,  Cor.  (&).) 

That  is,  if  A,  B,  and  the  third  point  are  not  in  the  same 
straight  line,  planes  FC  and  EG  coincide. 

But  this  conclusion  is  contrary  to  the  hypothesis. 

.'.  AB  is  a  straight  line. 

^  Q.E.D. 

Ex.  1.  To  make  sure  that  a  surface  is  perfectly  "flat,"  a  mechanic 
applies  his  "  straightedge  "  to  the  surface  in  various  directions  and  sees 
that  the  "straightedge"  touches  the  surface  along  its  whole  length  in 
every  position.     On  what  definition  is  his  action  based  ? 

Ex.  2.  How  may  three  points  be  so  situated  that  more  than  one  plane 
may  be  passed  through  them  ? 

Ex.  3.  Show  that  four  different  planes  may  be  passed  so  as  to  contain 
three  out  of  four  given  points,  if  no  three  of  the  points  be  collinear.  In 
how  many  planes  would  each  of  the  four  given  points  lie  ? 


THE  ELEMENTS  OF  GEOMETRY 


XXI.  2.  If  through  a  given  point  a  perpendicular  is 
drawn  to  a  plane,  it  is  the  only  perpendicular  that  can 
be  drawn  through  the  point  to  the  plane. 


p 

R/ 

^^^Fm 

/-^ 

\^^m    / 

Z^ 

■^■^1 

Hyp.    Case  I.     If  0  is  a  given  point  in  plane  JfQ,  and  PO 
is  perpendicular  to  plane  3fQ, 

Cone. :  then  PO  is  the  only  perpendicular  that  can  be  drawn 
to  plane  MQ,  at  0. 

Dem.     If  possible,  let  TtO  be  a  second  ±  to  MQ,  at  0. 
Let  the  plane  of  PO  and  RO  intersect  the  plane  MQ,  in  OA. 
Then  RO  must  be  ±  to  OA.  (Def.  of  JL  to  a  plane.) 

But  PO  is  perpendicular  to  OA.  (Def.  of  ±  to  a  plane.) 

.'.  we  have  OA  in  plane  of  PO  and  RO  perpendicular  to  PO 

and  RO. 

But  this  conclusion  is  impossible.  (Ax,  7.  Direct  Inf.) 

.-.  PO  is  the  only  perpendicular  that  can  bo  drawn  to  plane 

Mq  at  0. 

Q.E.D. 


Ex.  4.   Hold  two  pencils  so  as  to  show  that  if  two  lines  do  not  intersect, 
and  are  not  parallel,  a  plane  cannot  contain  both  of  them. 


XXI.     THE   PLANE   AND   ITS   RELATED  LINES        237 

O 

Hyp.     Case  II.     If  \ 

0    is    a    given    point  jH- 

without  the  plane  MQ, 
and  PO  is  perpen- 
dicular to  plane  MQ, 


Cone. :  then  PO  is  the  only  perpendicular  that  can  be  drawn 
to  plane  MQ  through  0. 

Dem.  If  possible,  let  OL  be  a  second  perpendicular  to  MQ 
through  0. 

Draw  LP. 

Then,  in  the  plane  OPLj  we  have  OP  and  OL  each  perpen- 
dicular to  PC. 

But  this  conclusion  is  impossible.  (Ax.  7.  Direct  Inf.) 

.-.  PO  is  the  only  perpendicular  that  can  be  drawn  to 
MQ  through  0.  ^^^ 

Ex.  5.  Why  are  the  projections  of  straight  lines  on  a  plane  always 
straight  ? 

Show  that  if  the  projection  of  a  line  on  a  plane  is  straight,  the  line 
need  not  be  straight. 

Ex.  6.  Show  how  a  circle  may  be  so  situated  with  respect  to  a  plane 
that  its  projection  on  the  plane  will  be  a  straight  line. 

Ex.  7.  Show  that  if  the  projection  of  the  line  AB  on  each  of  two  inter- 
secting planes  be  straight,  the  line  itself  must  be  straight. 

Ex.  8.  Why  does  a  three-legged  stool  always  stand  firmly  on  a  level 
floor,  while  a  table  or  chair  with  four  legs  may  be  unsteady  ? 

Ex.  9.  Show  that  if  two  lines  lie  in  the  same  plane,  they  must  either 
intersect  or  be  parallel. , 

Ex.  10.  Show  that  if  four  lines  concur,  the  greatest  number  of  planes 
that  can  be  determined  by  the  lines  two  and  two  is  six. 


288 


THE  ELEMENTS  OF  GEOMETRY 


XXI.  3.    A  line  perpendicular  to  two  lines  at  their 
intersection  is  perpendicular  to  the  plane  of  these  lines, 

A 


^^ 

yii' 

-~— ..NjV 

^y^ 

B                   M^ 

0 

/ 
/ 

\    \ 

/ 

\    V 

/ 

\. 

/ 

1 

A' 

Hyp.  If  AB  is  perpendicular  to  BM  and  BL  at  B,  and  if 
OK  is  the  plane  determined  by  BM  and  BL,  and  if  BR  is  any 
other  line  of  OK  through  B, 

Cone. :  then  AB  ±  BE,  or  AB  is  perpendicular  to  plane  OK. 

Dem.     Draw  EC,  any  line  of  OK 

Suppose  EC  cuts  BL  in  C,  BR  in  H,  and  BM  in  E. 

Produce  AB  to  A\  making  BA'  =  AB. 

Draw  AC,  AH,  AE,  and  A'C,  A'H,  A'E. 

EB  and  CB  are  mid-perpendiculars  to  AA'.  (Const.) 

.-.  EA  =  EA',  CA  =  CA'. 
EG  is  common  to  the  A  AEC,  A' EC. 

.-.  AAEC^AA'EC.  (V.  3.) 

.-.  ZACH=ZHCA'.       (Rom.  A  of  ^  A.) 

.'.  AACH^AHCA'.  (V.  1.) 

.-.  HA  =  HA',       (Hom.  sides  of  ^  A.) 

.'.  two  points  (B  and  H)  of  BR  are  equidistant  from  the 
ends  of  AA'. 

.:  EB  is  a  mid-perpendicular  to  AA'. 

.',  AB  is  perpendicular  to  any  line  of  OK  passing  through  B. 

.-.  AB  ±  plane  OK. 

^  Q.E.D. 


XXI.     THE   PLANE  AND   ITS   RELATED  LINES 


XXI.  3  a.  If  three  lines  are  perpendicular  to  a  given 
line  at  a  given  point,  these  perpendiculars  lie  in  a  plane 
perpendicular  to  the  given  line  at  the  given  point. 

XXI.  3  6.  The  plane  mid-normal  to  the  join  of  two 
points  is  the  locus  of  all  points  equidistant  from  the 
given  points. 

XXI.  3  c.  The  plane  through  a  given  point  perpen- 
dicular to  a  given  line,  is  unique,  whether  the  given 
point  he  on  the  given  line  or  ivithout  the  given  line. 

XXI.  4.  Of  all  straight  lines  drawn  from  a  given 
point  to  a  given  plane, 

(1)  The  perpendicular  is  the  shortest  line,  and  con- 
versely. 


Hyp.     If  PO  is  perpendicular  to  plane  MQ,  and  PA  is  any 

other  line  from  P  to  plane  MQ, 

Cone:  then  PO<PA. 

Dem.     Draw  AO. 

In  rt.  A  POA,  PO  <  PA,  (IV.  4  a.  Sch.) 

.-.  PO<PA. 

Q.E.D. 

Proof  of  converse  is  left  as  an  exercise  for  the  pupil. 


240 


THE  ELEMENTS  OF  GEOMETRY 


(2)   Obliques  with  equal  projections  are  equal,  and 
conversely. 


Hyp.  liAR  and  ^(7  are  obliques  drawn  from  the  point  A  to  the 
plane  MQ,  with  the  equal  projections,  BR  and  BCj  respectively, 

Cone:  then  AR  =  AC. 

Dem.  Rt.  A  ABR  ^  rt.  A  ABC.  (V.  1.) 

.-.  AR  =  AC.         (Horn,  sides  of  ^  A.) 

Q.E.D. 

Proof  of  converse  is  left  as  an  exercise  for  the  pupil. 

(3)  Of  two  obliques  ivith  unequal  projections, '  that 
one  is  the  greater  which  has  the  greater  projection,  and 
conversely. 

Hyp.     If  projection  BF  is  greater  than  projection  BR, 
Cone:  then  AF>AR. 

Dem.     Lay  off  -80=  BR,  and  draw  AC 

AC=AR.  (Why?) 

But  AF>AC. 

.-.  AF>AR. 


Proof  of  converse  is  left  as  an  exercise  for  the  pupil. 


Q.E.D. 


XXI.     THE   PLANE    AND   ITS    liELATED   LINES 


Theorem  of  the  Three  Perpendiculars 


241 


XXI.  5.  If  from  the  foot  of  a  perpendicular  to  a 
plane  a  line  is  drawn  at  right  angles  to  a  second  liiie 
of  the  plane,  and  if  the  point  of  intersection  of  these 
two  lines  is  joined  to  any  point  in  the  perpendicidar, 
this  third  line  is  perpendicular  to  the  second  line  of  the 
plane. 


Hyp.  If  AB  is  perpendicular  to  plane  MQ,  and  if  BE  in  MQ 
is  drawn  from  the  foot  of  AB  perpendicular  to  LF,  any  line 
of  MQ,  and  if  from  E,  the  point  of  intersection  of  BE  and  LF, 
AE  is  drawn  to  A,  any  point  in  AB, 

Cone:  then  AE  ±  FL. 

Dem.     Take  EU  =  EC.     Draw  BR,  BC,  AE,  and  AC. 

ABEC^ABEB.  (V.  1.) 

.-.  BC==BR.        (Horn,  sides  of  ^  A.) 

.-.  AC=AB.   ■  (XXI.  4.  (2).) 

,\AE±FL.  (VIII.  2.  Sch.  2.) 

Q.E.D. 

XXI.  5  a.  The  second  line  FL  in  the  above  figure  is 
perpendicular  to  the  plane  of  the  first  and  third  lines, 
namely,  plane  AEB. 


^  .         ^ 


242 


THE  ELEMENTS  OF  GEOMETRY 


XXI.  5  b.  If  two  lines  intersect  at  right  angles,  and 
through  their  point  of  intersection  a  perpendicular  to 
the  second  is  drawn  without  the  plane  of  the  lines,  the 
first  is  the  projection  of  the  third  perpendicular  on  the 
plane  of  the  given  lines. 

XXI.  6.  If  one  of  two  parallels  is  perpendicular  to  a 
plane,  the  other  is  also  perpendicular  to  the  plane,  and 
conversely.  » 


M 


/E 


G' 


Hyp.     If  AB  II  CE^  and  AB  is  perpendicular  to  plane  MQ, 

Gone. :  then  CE  is  perpendicular  to  plane  MQ. 

Dem.     Draw  BE\    also,   in   plane   MQ,   GF  ±  BE  at  E; 
draw  AE. 

GF  is  perpendicular  to  plane  of  AE  and  EB.       (XXI.  5  a.) 
But  CE  lies  in  plane  EAB.  (XXI.  Def.  of  plane,  (e).) 

.-.  GE±  CE.  (Def.  J  to  a  plane.) 

.-.  CE±GE. 

But  CE±  BE.        (Def.  lis,  2d  Dir.  Inf.) 

.-.  CE±  plane  MQ.  (XXI.  3.) 

Q.E.D. 
Conversely.  If  AB  and  CE  are  perpendicular  to  plane  MQ, 
Cone. :  then  AB  II  CE. 


XXI.     THE   PLANE   AND   ITS   RELATED   LINES         243 


Dem.     If  EC  is  not  parallel  to  BA,  draw  ET  that  is. 
Then  ^T  is  perpendicular  to  plane  MQ.  (XXI.  6.) 

But  EC  is  perpendicular  to  plane  MQ,  (Hyp.) 

And  EC  is  the  only  perpendicular  that  can  be  drawn  to 
plane  MQ  at  E.  (XXI.  2.) 

.*.  ET  and  EG  must  coincide. 


.-.  AB  II  GE. 


Q.E.D. 


XXI.  6  a.    If  two  lines  are  parallel  to  a  third,  they 
are  pai^allel  to  each  other. 


Hyp.     If  AB  II  GR 
and  EF II  CRy 


Cone:  then 


A 

E 

G 

M 

■ 

\ 

■ 

^ 

A 

AB  II  EF. 


Dem.     Pass  a  plane  MQ  perpendicular  to  GR. 

Then  AB  and  ^i^  are  each  perpendicular  to  MQ.     (XXI.  6.) 

.-.  ^5  II  EF.  (XXI.  6.  Conv.) 

Q.E.D. 

Ex.  11.  A  ruled  surface  is  one  that  may  be  generated  by  the  motion 
of  a  straight  line.     Show  that  a  plane  is  a  ruled  surface. 

Ex.  12.  To  get  a  "straightedge"  we  sometimes  fold  a  sheet  of  paper 
and  use  the  edge  of  the  fold.    What  proposition  are  we  illustrating  ? 

Ex.  13.  In  how  many  positions  can  the  pendulum  of  a  clock  be  per- 
pendicular to  the  floor  on  which  the  clock  stands  ?     Why  ? 

Ex.  14.  To  find  whether  or  not  a  square  post  is  perpendicular  to  a 
floor,  a  carpenter  applies  a  square  to  the  floor  and  post  on  two  sides  of 
the  post.  On  what  theorem  is  his  action  based?  Does  it  make  any 
difference  what  sides  of  the  post  he  selects  ?    Why  ? 

Ex.  15.  To  keep  a  vertical  sign  in  an  upright  position,  it  is  fastened  at 
the  foot  to  two  horizontal  crosspieces  nailed  together  in  the  shape  of  an 
X.     What  proposition  is  illustrated  by  this  device  ? 


244  THE   ELEMENTS  OF   GEOMETRY 

XXI.  7.  If  two  angles  not  in  the  same  plane  have 
their  sides  respectively  parallel  and  lying  on  the  same 
side  of  the  join  of  their  vertices,  the  angles  are  equal. 


Hyp.     If  A'V  II  AL  and  A'R  II  AR,  and  these  lines  lie  on 
same  side  of  the  join  AA^  and  also  lie  in  different  planes, 

Cone:  then  ZL'A'E' =  ZLAR. 

Dem.     Lay  off  AB=A'B',  and  AO=A'a';  and  join  CO,  BB\ 
BO,  and  B' v. 
The  4-sides  A-B'  and  A-C  are  parallelograms.  (VI.  2.) 

.-.  BB'  =  AA'  and  CO  =  AA'.  (VI.  1  a.) 

.-.  BB'=Ca.  (Ax.  1.) 

BB'  II  CO,  (XXI.  6  a.) 

••.  the  4-side  B-C  is  a  parallelogram.         (VI.  2.) 

.-.  BC=B'a. 

.-.  A  JBC^  ^  A  B'OA'.  (V.  3.) 

.-.  Z  L'A'R'  =  Z  X^iJ.    (Hom.  Aof  ^  A.) 

Q.E.D. 


XXI.     THE   PLANE   AND   ITS   RELATED  LINES        245 

XXI.  %.  If  a  line  is  parallel  to  a  plane,  any  parallel 
to  the  given  line  through  a  point  of  the  plane  lies  wholly 
in  the  plane. 


Hyp.     If  AB  is  parallel  to  plane  MQ,  and  CE  II  AB  through 
G,  a  point  of  MQ, 

Cone. :  then  CE  lies  wholly  in  the  plane  MQ. 

Dem.     But  one  parallel  to  AB  can  pass  through  C.     (Ax.  7.) 
The  line  of  intersection  of  the  planes  ABC  and  MQ  is  paral- 
lel to  AB  and  passes  through  C.  (Def.  of  II  to  a  plane.) 
.-.  this  line  of  intersection  is  the  parallel  CE,  that  is,  CE 
lies  wholly  in  the  plane  MQ. 


Q.E.D. 


Ex.  16.  How  many  lines  may  be  drawn  perpendicular  to  a  given  line 
at  a  given  point  ?     How  are  all  these  lines  situated  ? 

Ex.  17.  How  many  lines  can  be  drawn  perpendicular  to  a  given  line 
through  a  point  without  this  line  ?    Prove. 

Ex.  18.  Prove  that  one  plane,  and  only  one,  may  be  passed  perpen- 
dicular to  a  given  line  at  a  given  point. 

Ex.  19.  Show  how  to  pass  a  plane  through  a  given  point  perpendicular 
to  a  given  line  passing  through  the  point. 

Ex.  20.  Show  how  to  pass  a  plane  through  a  given  point  perpendicular 
to  a  line  that  does  not  pass  through  the  point. 

Ex.  21.  What  surface  is  generated  by  the  hand  of  a  clock  as  it  passes 
around  the  dial  ?     Why  ? 

Ex.  22.  A  vertical  flagstaff  75  ft.  high  stands  in  the  center  of  a  grass- 
plot  40  ft.  in  diameter.  How  far  is  the  top  of  the  pole  from  any  pohit  in 
the  circumference  of  the  grassplot  ? 

Ex.  23.  The  blades  of  a  windmill  are  15  ft.  long,  and  are  fastened  to 
the  axle  at  an  angle  of  60°.  How  far  does  the  tip  of  a  blade  travel  in  10 
minutes,  when  the  wheel  is  making  80  revolutions  a  minute  ? 


246         THE  ELEMENTS  OF  GEOMETRY 

XXI.  8  a.  Conversely.  If  a  line  is  parallel  to  a  line 
of  a  plane f  it  is  parallel  to  the  plane. 

A B 

Hyp.   UABWCEy  M   I 

a  line  of  the  plane  /  2>L= 'i?^ 

MQ,  /  c 

Cone. :  then  AB  is  parallel  to  plane  MQ. 

Dem.     Through  AB  pass  any  plane  AR. 
Let  this  plane  intersect  MQ  in  DR. 

If  DR  -^  AB,  DR  ^  CE.  (XXI.  6  a.) 

If  DR^  CE,  draw  DS  II  CE. 

DS  lies  in  MQ.       (Def.  of  plane,  (d).) 

DSWAB.  (XXI.  6  a.) 

*.  DS  is  coplanar  with  AB.  (Def.  of  plane,  (d).) 

But  DR  is  coplanar  with  AB.  (Const.) 

.-.  the  two  planes  ABDR  and  ABDS  have  three  points  {A, 
B,  D)  common. 

But  this  is  absurd.  (Def.  of  plane,  (6).) 

.-.  to  suppose  DR  -^  AB  is  absurd. 

.-.  DR  II  AB. 

That  is,  AB  is  parallel  to  plane  MQ.        (Def.  of  II  to  plane.) 

Q.E.D. 

Ex.  24.  The  arm  of  a  derrick  is  50  ft.  long ;  it  is  so  fastened  to  the 
mast  as  to  revolve  at  a  constant  angle  of  30°  with  the  vertical  upright. 
How  far  does  the  end  of  the  arm  travel  in  a  quarter  revolution  ? 

Ex.  25.  The  ceiling  of  a  room  is  10  ft.  high.  How  would  you  deter- 
mine, by  means  of  a  12-ft.  pole,  a  point  in  the  floor  directly  under  a  gas 
drop  in  the  ceiling  ? 

Ex.  26.  If  two  columns  are  perpendicular  to  the  same  floor,  how  are 
they  situated  with  respect  to  each  other  ? 


XXI.     THE   PLANE   AND   ITS   RELATED   LINES        247 

XXI.  S  b.  If  a  line  is  parallel  to  each  of  two  planes, 
it  is  parallel  to  their  line  of  intersection. 

Hyp.     If  AB  is  II  to 

plane  EF,  and  also  to 

plane  EG, 

G 

Cone. :  then  AB  II  EC. 

Dem,     A  parallel  to  AB  through  C  must  lie  in  the  plane  EF. 

(XXI.  8.) 

This  parallel  must  also  lie  in  the  plane  EG.  (XXI.  8.) 

.-.  the  parallel  must  be  CE. 

^  Q.E.D. 

Ex.  27.  A  column  is  perpendicular  to  a  level  floor.  The  capital  and 
base  of  a  second  column  in  the  same  wall  are  respectively  20  ft,  from  the 
capital  and  base  of  the  first.  By  what  proposition  do  you  know  the  second 
column  to  be  vertical  ? 

Ex.  28.  How  would  you  make  use  of  XXL  5  6  to  let  fall  a  perpen- 
dicular to  a  plane  from  a  point  without  the  plane  ? 

Ex.  29.  Show  how  to  draw,  through  a  given  point  in  a  plane,  a  per- 
pendicular to  the  plane  by  using  XXI.  5  6. 

Ex.  30.  Show  how  to  pass  a  plane  through  a  given  point  parallel  to  a 
given  line. 

Ex.  31.  How  many  planes  may  be  passed  through  a  given  point  parallel 
to  a  given  line  ? 

Ex.  32.  Show  how  to  draw  a  line  through  a  given  point  parallel  to  a 
given  plane. 

Ex.  33.  How  many  lines  may  be  drawn  through  a  given  point  parallel 
to  a  given  plane  ? 

Ex.  34.  If  a  number  of  lines  be  drawn  through  a  given  point  parallel 
to  a  given  plane,  how  will  these  lines  be  situated  ?    Why  ? 

Ex.  35.  Show  that  if  a  number  of  lines  be  parallel  to  the  same  plane, 
and  one  of  the  lines  intersect  all  the  others,  then  all  the  lines  must  lie  in 
the  same  plane. 

Ex.  36.  If  a  number  of  planes  be  passed  through  a  given  point  parallel 
to  a  given  line,  how  will  these  planes  b^  situated  With  respect  to  each 
other  ?    Why  ? 


248  THE  ELEMENTS  OF  GEOMETRY 

XXI.     SUMMARY   OF  PROPOSITIONS   IN   THK  GROUP 
ON  THE   PLANE  AND   ITS   RELATED   LINES 

1.  The  line  of  intersection  of  two  planes  is  straight 

2.  If  through  a  given  point  a  perpendicular  is  draivn 
to  a  given  plane,  it  is  the  only  one  that  caii  he  drawn 
through  the  point  to  the  plane. 

Case  I.    Point  in  plane. 

Case  II.    Point  without  plane, 

3.  A  line  perpendicular  to  each  of  two  other  lines  at 
their  intersection  is  perpendicular  to  the  plane  of  these 
lines, 

a.  If  three  lines  are  perpendicular  to  a  given  line 
at  a  given  point,  these  perpendiculars  lie  in 
a  plane  perpendicular  to  the  given  line  at 
the  given  point. 

h.  The  plane  mid-normal  to  the  join  of  two  points 
is  the  locus  of  all  points  equidistant  from 
the  given  points. 

c.  The  plane  through  a  given  point  perpendicular 
to  a  given  line,  is  unique,  ivhether  the  given 
point  he  on  the  given  line  or  without  the 
given  line. 

4.  Of  all  straight  lines  drawn  from  a  given  point  to 
a  given  plane, 

(1)  The  perpendicular  is  the  shortest  line,  and  con- 
versely. 


XXI.     THE   PLANE   AND   ITS  RELATED   LINES        251 
PROBLEMS 

XXI.  Prob.  1.     Through  a   given  point  to  draw  a 
perpendicular  to  a  given  plane. 


A 

\\ 

I    \ 

M !        \  n  M. 


F 


B  Q  B 

Case  L  Case  II. 

Given.     The  plane  MQ  and  the  point  A. 
Required.     A  perpendicular  to  MQ  through  A. 

Case  I.     A  is  without  MQ. 

Const.     Draw  any  line,  BG,  in  MQ. 

Draw^E±£a 

On  MQ  draw  EF  A.  BO. 

Dt^w  AG±EF. 

AG  is  the  perpendicular  required. 

Dem.     The  demonstration  is  supplied  by  XXI.  5  b. 

Case  II.    J.  is  in  MQ. 

Const.     Draw  any  line,  AE,  in  MQ. 

In  MQ  dvsiw  BG±AE. 

Draw  EF,  without  MQ,  and  perpendicular  to  BG. 

In  the  plane  of  FE  and  AE  draw  AG  ±  AE. 

AG  is  the  perpendicular  required. 

» Dem.     The  demonstration  is  supplied  by  XXI.  5  a. 


Q.E.P. 


Q.E.F. 


252         THE  ELEMENTS  OF  GEOMETRY 

XXI.  Prob.  2.   To  draw  a  common  perpendicular  to 
two  lines  not  in  the  same  plane. 


Q 
Given.     The  lines  AB  and  CE,  not  in  the  same  plane. 

Required.     A  common  perpendicular  to  AB  and  CE. 

Const.     Through  any  point  of  CE,  as  F,  draw  FO  II  AB. 
CE  and  FO  determine  a  plane  MQ.  (Def.  of  plane,  (a).) 

Project  BA  on  this  plane,  by  the  perpendiculars  HL,  AS. 
Let  CE  intersect  SL  in  0. 

Through  0  draw  0J±  to  the  plane  MQ.       (XXI.  Prob.  1.) 
OJ  is  the  perpendicular  required. 

Dem.                            AB  II  MQ.  (XXI.  8  a.) 

AB  li  SL.  (Def.  of  II  to  a  plane.) 

OJJL  SL.  (Def.  of  ±  to  plane.) 

OJ  II  AS.  (XXI.  6,  converse.) 

.'.  OJ  lies  in  the  plane  through  0,  xS",  and  ^ ; 
i.e.  in  the  plane  of  the  parallels  AB  and  SL. 
.'.  OJ  intersects  AB,  and  OJAS  is  a  rectangle.     (Def.  of  □.) 

i.e.  0J±  AB. 

But  0J±  CE.      (Def.  of  ±  to  a  plane.) 

.*.  OJ  is  a  common  perpendicular  to  ^IB  and  CE. 

^     ^  Q.E.D. 

ScH.  But  one  common  perpendicular  can  be  drawn  to  AB 
and  CE ;  for  a  line  perpendicular  to  AB  must  be  perpendicu- 
lar to  SLf  which  is  parallel  to  ^B.  A  common  perpendicular 
to  AB  and  CE  must  therefore  be  perpendicular  to  CE  and  SL, 
and  hence  perpendicular  to  MQ  at  0.  But  the  line  perpendic- 
ular to  MQ  at  0  is  unique  j  that  is,  the  common  perpendicular 
is  unique. 


XXII.     GROUP   ON   PLANAL   ANGLES 

DEFINITIONS 

(a)    Dihedrals 

A  Dihedral  angle,  or  simply  a  dihedral,  is  the  figure  formed 
by  two  planes  that  intersect. 

The  Faces  of  a  dihedral  are  the  planes 
by  which  it  is  formed  (AE,  FG).  f  G 

The  Edge  of  a  dihedral  is  the  line  of  intersection  of  the  faces 
{CE). 

When  two  planes  intersect  so  that  the  four  dihedrals  formed 
are  equal,  each  of  the  dihedrals  is  called  a  Right  Dihedral,  and 
the  planes  are  said  to  be  perpendicular  to  each  other. 

The  terms  acute,  obtuse,  oblique,  angles  of  the  same  kind,  com- 
plemental,  supplemental,  adjacent,  opposite,  exterior,  interior,  corre- 
sponding, alternate,  etc.,  have  the  same  meaning  in  solid  geometry 
as  in  the  plane,  the  face  of  the  dihedral  replacing  the  side  of 
the  plane  angle. 

Method  of  reading  Dihedrals.  A  dihedral  may  be  read,  when 
there  is  no  ambiguity,  by  merely  reading  the  edge;  as, 

dihedral  CE. 

Otherwise,  the  angle  is  indicated  by  reading  one  letter  from 
each  face,  with  the  edge  between  these  letters ;  thus, 

A-CE-G. 

The  dihedral  may  be  assumed  to  be  generated  by  the  revolu- 
tion of  a  plane,  upon  the  edge  as  an  axis,  from  an  initial  plane 
(as  FG)  to  a  terminal  plane  (as  AE). 

As  in  plane  geometry,  so  in  solid  geometry,  rotation  is  Posi 
tive  when  anti-clockwise,  and  Negative  when  clockwise. 

253 


254  THE  ELEMENTS  OF  GEOMETRY 

The  Rectilineal  Angle  of  a  dihedral  is  the  plane  angle  formed 
by  two  lines,  one  drawn  in  each  face,  and  perpendicular  to  the 
edge  at  the  same  point. 

Note.  —  It  is  easily  shown  that 

(a)  If  two  dihedrals  are  equal,  their  rectilinear  angles  are  equal,  and 
conversely. 

(6)  Any  two  dihedrals  are  to  each  other  a.s  their  rectilinear  angles. 

Accordingly,  the  rectilineal  angle  is  usually  called  the  Measure  of  the 
dihedral. 

A  Transversal  Plane  is  a  plane  intersecting  a  number  of  other 
planes. 

Parallel  Planes.  Two  planes  are  said  to  be  parallel  when 
they  are  so  situated  that  if  any  transversal  plane  cuts  them, 
the  corresponding  exterior-interior  dihedrals  are  equal  and 
have  their  edges  parallel. 

The  plane  through  a  given  point  and  parallel  to  a  given 
plane  is  unique. 

(6)     POLYHEDRALS 

A  Polyhedral  (n-dral)  is  the  figure  formed  by  the  intersection 
of  three  or  more  planes  at  a  single  point. 

Dihedral  and  polyhedral  angles  are  called,  collectively, 
Planal  Angles. 


The  Vertex  of  the  polyhedral  is  the  point  in  which  the  planes 
intersect  (0). 

The  Faces  of  a  polyhedral  are  the  intersecting  planes  (AOB, 
BOC,  etc.). 

The  Edges  of  a  polyhedral  are  the  lines  of  intersection  of  the 
faces  (0-4,  OB,  etc.). 


XXII.     PLANAL   ANGLES  255 

The  Face  Angles  of  a  polyhedral  are  the  plane  angles  formed 
at  the  vertex  by  consecutive  edges  {Z.AOC,  Z  BOC,  etc.). 

Symmetric  Polyhedrals.  As  the  faces  and  edges  of  a  poly- 
hedral are  infinite  (unlimited)  in  extent,  the  polyhedral  will 
have  two  parts,  lying  on  opposite  sides  of  the  vertex,  the  angles 
of  which,  dihedral  and  plane,  are  equal  in  pairs. 

Thus,  B-AO-C  =  B'-A'O-C. 

(Their  faces  are  the  same  planes.) 

Similarly,  A-BO-C  =  A'-B'O-C,  etc. 

Again,  Z  AOC  =  Z  A'OC   (vertical  angles),  etc. 

But  the  equal  angles  of  the  two  parts  of  the  polyhedral  occur 
in  reverse  order,  as  indicated  by  the  arrowheads. 

For  this  reason  the  two  parts  of  the  polyhedral  are  called 
Symmetrical  Polyhedrals.     This  name  is  due  to  Legendre. 

Unless  the  contrary  is  stated,  or  evidently  implied,  we  shall, 
in  using  the  word  polyhedral,  refer  to  the  part  on  one  side  of 
the  vertex  only.     The  other  part  will  be  called  the  symmetrical. 

Polyhedrals  are  classified  according  to  the  number  of  faces 
as  tri(3)hedrals,  tetra(4)hedrals,  penta(5)hedrals,  etc. 

The  Trihedral  is  analogous  to  the  triangle,  the  faces  of  the 
trihedral  corresponding  to  the  sides  of  the  triangle,  and  the 
dihedrals  of  the  trihedral  to  the  angles  of  the  triangle. 

Hence,  the  following  terms  will  be  self-explanatory : 

Scalene,  Isoangular, 

Isosceles,  Equiangular. 

Equilateral, 

A  Rectangular  trihedral  is  one  that  has  a  single  right  dihedral. 

A  Bi-rectangular  trihedral  is  one  that  has  two  right  dihedrals, 
and  no  more. 

A  Tri-rectangular  trihedral  is  one  that  has  three  right  dihe- 
drals. 

Method  of  reading  Polyhedrals.  A  polyhedral  is  read  by 
taking  first  the  letter  at  the  vertex,  and  then,  in  succession, 
the  letters  at  the  other  extremities  of  the  edges,  as  0-ABCE. 


256 


THE  ELEMENTS  OF  GEOMETRY 


If  no  misunderstanding  is  likely  to  result,  the  letter  at  the 
vertex  may  be  used  alone,  as  in  the  case  of  plane  angles. 

A  Convex  Polyhedral  is  one  whose  intersection  with  any  plane 
not  passing  througli  its  vertex  is  a  convex  polygon. 


PROPOSITIONS 


(a)  Dihedrals 

XXII.  1.  In  a  right  dihedral  a  line  drawn  in  one 
face,  perpendicular  to  the  edge,  is  perpendicular  to  the 
other  face. 


Hyp.     If  P-SO-Q  is  a  right  dihedral,  and  if  AB  of  plane  PO 

is  perpendicular  to  SO,  the  edge  of  the  dihedral, 

Cone. :  then  ^B  is  perpendicular  to  plane  MQ. 

Dem.     Through  A  draw  CR  in  plane  MQ,  perpendicular  to 

SO. 

Z  CAB  is  the  measure  of  the  dihedral  P-SO-Q. 

(Def.  of  measure  of  a  dihedral.) 

.*.  Z  OAB  is  a  right  angle. 

But  AB±SO.  (Hyp.) 

.-.  AB  is  perpendicular  to  plane  MQ.        (XXI.  3.) 

Q.E.D. 


XXII.     PLANAL  ANGLES 


257 


XXII.  1  a.    A  perpendicular  to  either  face  of  a  right 
dihedral,  at  any  point  of  the  edge,  lies  in  the  other  face. 

Hyp.     If  AB  is  perpendicular  to  plane  MQ,  one  face  of  the 
dihedral  F-SO-Q, 

Cone. :  then  AB  lies  in  plane  OP. 

Dem.     A  line  in  plane  OP  perpendicular  to  SO  is  perpendicu- 
lar to  plane  MQ.  (XXII.  1.) 
Only  one  _L  to  plane  MQ  can  be  drawn  at  J..  (XXI.  2.) 
But  plane  OP  is  perpendicular  to  plane  MQ.  (Hyp.) 


.-.  AB  must  lie  in  plane  OP. 


Q.E.D. 


Ex.  1.  If  a  line  and  a  plane  are  each  perpen- 
dicular to  a  second  plane,  the  line  and  plane  are 
parallel. 


Ex.  2.  If  a  line  a  be  perpendicular  to  a  plane 
MQ,  and  the  lines  b  and  c  be  both  perpendicular 
to  a,  then  6  and  c  will  both  be  parallel  to  plane 


J 


M 


<7 


Ex.  3.  If  two  intersecting  lines  be  parallel  to  a 
given  plane,  the  plane  of  the  lines  will  be  parallel 
to  the  given  plane. 

Ex.  4.  Hence,  show  how  to  pass,  through  a 
given  point,  a  plane  parallel  to  a  given  plane. 


258  THE   ELEMENTS  OF  GEOMETRY 

XXII.  2.  If  each  of  two  intefi^secting  i^lanes  is  per- 
pendicular to  a  third,  their  line  of  intersection  is  per- 
pendicular to  the  third. 


Hyp.  If  plane  CE  is  perpendicular  to  plane  MQ,  and  plane 
FG  is  perpendicular  to  plane  MQ,  and  AB  is  their  line  of 
intersection, 

Cone. :  then  AB  is  perpendicular  to  plane  MQ, 

Dem.    Dihedral  AG  is  a.  right  dihedral.  (Hyp.) 

.-.  a  ±  to  plane  MQ  at  A  lies  in  plane  FG.  (XXII:  1  a.) 

Also  a  X  to  plane  MQ  at  A  lies  in  plane  CE.  (XXII.  la.) 

.-.  AB  is  perpendicular  to  plane  MQ. 

Q.E.D. 


Ex.  5.   If  one  of  two  parallels  is  parallel  to  a 
given  plane  MQ,  the  other  is  also  parallel  to  MQ.  M 

Ex.  6.    If  each  of  two  planes  is  parallel  to  a 
third,  these  two  planes  are  parallel  to  each  other. 

Ex.  7.   Through  a  given  point  but  one  plane  can 
be  passed  parallel  to  two  lines  that  are  not  parallel  to  each  other. 


ZZ7 


XXII.     PLANAL   ANGLES  259 

XXII.  2  a.  If  a  line  is  per^Jendicular  to  a  plane, 
every  plane  through  the  line  is  also  perpendicular  to 
the  plane.  gc 

Hyp.     If  AB  is 

perpendicular     to 

plane     MQ,     and  jf^ 

plane  GF  is  drawn  / 

through  ABj  / 


^ 


F  Q 

Cone. :  then  plane  GF  is  perpendicular  to  plane  MQ. 

Dem.    Draw  AJS±  CF  and  in  plane  MQ. 

AB  ±  AE.        (Def .  of  ±  to  a  plane.) 

AE1.CF.  (Const.) 

.•.  AE  is  perpendicular  to  plane  GF.       (XXI.  3.) 

But  rt.  Z  EAB  is  the  measure  of  dihedral  Q-FC-G. 

,\  plane  GF  is  perpendicular  to  plane  MQ. 

Q.E.D. 

Ex.  8.  What  is  the  smallest  number  of  plane  angles 
that  can  be  brought  together  at  a  point  to  form  a  convex 
polyhedral  ?  What  must  be  true  of  the  sum  of  these 
angles  if  they  form  a  polyhedral  ? 

Ex.  9.    If  three  equilateral  triangles  be  brought  to- 
gether so  as  to  have  a  common  vertex,  what  will  be  the 
sum  of  the  plane  angles  at  this  vertex  ?    Why,  then,  must  a  polyhedral 
be  formed  ? 

Ex.  10.    Show  that  a  polyhedral  may  be  formed  with  four  equilateral 
triangles  ;  also  with  five. 

Ex.  11.    Show  that  no  polyhedral  can  be  formed  by  using  any  larger 
number  of  equilateral  triangles  than  five. 

Ex.  12.    Show  that  if  all  the  faces  of  a  polyhedral  be  regular  w-gons, 
the  only  polyhedrals  possible  are  those  in  which  w  =  3,  4,  or  5. 


260         THE  ELEMENTS  OF  GEOMETRY 

(h)    TOLYHEDRALS 

XXIT.  3.    ITie  sum  of  the  two  face  angles  of  a  tri- 
hedral is  greater  than  the  third  angle. 

Note.  —  No  proof  is  necessary  unless  the  third  angle  is  greater  thap 
each  of  the  others. 


Hyp.     If  F-ABC  is  a  trihedral  angle,  and  Z  AFC  is  greater 
than  either  Z  AFB  or  Z  BFC, 

Cone. :  then        Z  AFB  +  Z  BFC  >  Z  AFC 

Dem.     Draw  FR  iii  plane  AFC  so  that  Z  AFB  =  Z  AFB. 
Lay  off  FE  =  FB  and  pass  a  plane  through  B  and  B,  cutting 
the  edges  in  A,  B,  and  O. 
Draw  AB,  BC,  and  AC. 

A  AFB  ^  A  AFB.  (V.  1.) 

.-,  AB  =  AB. 

But  AB  +  BC>  AC.  (VIT.  1.) 

.-.  BC  >  BC.  (Preliminary  Th.  3.) 

.-.  ZBFC>ZBFC 

[If  two  A  have  two  sidee  of  one  equal,  etc.]       (VII.  Ex.  9.) 

.-.  ZAFB-{-ZBFC>ZAFB-hZBFC(  =  ZAFC). 

Q.E.D. 


XXII.     PLANAL  ANGLES 


261 


XXII.  4.    The  sum  of  the  face  angles  of  any  convex 
polyhedral  is  less  than  four  right  aiigles. 


Hyp.     If  i^  is  a  convex  polyhedral, 

Cone. :  then  Z  AFB  +  Z  BFG  +  Z  CFE,  etc.,  <  4  rt.  A. 

Dem.     Pass  a  plane  cutting  the  edges  of  polyhedral  F  in 
A,  B,  C,  .... 

From  0,  any  point  in  this  plane,  draw  OA,  OB,  00,  •••. 

In  the  base  there  are  n  triangles ;  there  are  also  n  triangle 
faces  in  solid  angle  F. 

.'.  the  sum  of  the  int.  A  of  the  base  A  =  2n  rt.  A,     (III.  1.) 
and  the  sum  of  the  int.  A  of  the  face  A  =  2  n  rt.  A      (III.  1.) 

Now    Z  GAB  of  the  base  <AFAG  +  A  FAB.      (XXII.  3.) 

Similarly,  Z  ABC<  Z  FBA  +  ZFBC.       (XXII.  3.) 

.-.  the  sum  of  the  angles  of  the  base  n-gon  is  less  than  the 
sum  of  the  base  angles  of  the  face  triangles. 

Now  the  sum  of  the  interior  angles  of  the 

n-gon  =  (2  71  -4)  rt.  Z ;  (III.  3.) 

that  is,  the  sum  of  the  angles  about  0  =  4  rt.  A. 

.-.  as  the  sum  of  the  base  angles  of  the  face  triangles  is  greater 

than  the  sum  of  the  interior  angles  of  the  n-gon,  it  follows  that 

ZAFB  +  ZBFC-i-ZGFE"'<4:Vt.A.  q.e.d. 


262         THE  ELEMENTS  OF  GEOMETRY 

XXII.  4.  ScH.  If  planal  angle  F  were  concave  in  any  of 
its  dihedrals,  the  sum  of  the  angles  about  angle  F  might  ex- 
ceed four  right  angles. 

XXII.  6.  A  line  perpendicular  to  one  of  two  parallel 
planes  is  perpendicular  to  the  other,  and  conversely. 


Hyp.  If 
plane  MQ 
is  parallel 
to  plane  RS 

and 
ABl.RSy 


a 

Cone. :  then  AB  ±  MQ. 

Dem.     Through  B  draw  any  two  lines,  BF  and  BG  in  RS. 
Let  planes  FBA  and  GBA  intersect  MQ  in  AC  and  AE, 
respectively. 
Then,  AC  II  BF  and  AE  II  BG.  (Def.  II  pis.) 

But,  AB±BF,  (Def.  ±  pi.) 

.-.  AB  ±  AC  (Def.  ll's.  Direct  Inf.  (2).) 
Similarly,  AB  A.  AE. 

.:  AB  ±  RS.  (XXI.  3.) 

Q.E.D. 
Conversely ; 

Hyp.     If  planes  MQ  and  RS  are  J_  AB, 

Cone. :  then  MQ  11  RS. 

Dem.     The  plane  through  A,  II  RS,  is  unique.     (Def.  II  pis.  a.) 
The  plane  through  A  ±  AB  is  unique.  (XXI.  3  c.) 

The  11  plane  is  ±  AB.  (Direct.  Th.) 

.•.  the  plane  ±  AB  is  the  II  plane.  (Ax.  10.) 

That  is,  MQ  II  RS.  ^  „  ^ 

'  ^  Q.E.D. 


XXII.     PLANAL  ANGLES  263 

XXII.      SUMMARY   OP    PROPOSITIONS   IN   THE   GROUP 
ON  PLANAL   ANGLES 

(a)   Dihedrals 

1.  In  a  right  dihedral  a  line  drawn  in  one  face,  per- 
pendicular  to  the  edge,  is  perpendicular  to  the  other 
face. 

a.  A  perpendicidar  to  either  face  of  a  right  dihe- 
dral, at  any  point  of  the  edge,  lies  in  the 
other  face. 

2.  If  each  of  two  intersecting  planes  is  perpendicular 
to  a  third,  their  line  of  intersection  is  perpendicular  to 
the  third. 

a.  If  a  line  is  perpendicular  to  a  plane,  every  plane 
through  the  line  is  perpendicular  to  the  given 
plane. 

(h)    POLYHEDRALS 

3.  The  sum  of  any  two  face  angles  of  a  trihedral  is 
greater  than  the  third  angle. 

4.  The  sum  of  the  face  angles  of  any  co7ivex  poly- 
hedral is  less  than  four  right  angles. 

ScH.  If  planal  angle  F  were  concave  in  any  of  its  dihedrals, 
the  sum  of  the  angles  about  angle  F  might  exceed  four  right 
angles. 

'  5.  A  line  perpendicular  to  one  of  two  parallel  planes 
is  perpendicular  to  the  other ^  and  conversely. 


264  THE  ELEMENTS  OF  geomp:try 

Show  that  if  the  faces  of  a  polyhedral  are  to  be  regular  polygons  of 
different  kinds,  it  is  always  possible  to  form  a  polyhedral  with  the  follow- 
ing combinations  of  figures : 

Ex.  13.   Two  equilateral  triangles,  with  any  regular  polygon  whatever. 

Ex.  14.   Three  equilateral  triangles,  with  any  regular  polygon  whatever. 

Ex.  15.  Two  squares,  with  any  regular  polygon  whatever. 

In  how  many  ways  can  polyhedrals  be  formed  by  using,  at  a  common 
vertex : 

Ex.  16.  Squares  ?  Ex.  17.   Regular  pentagons  ? 

Ex.  18.  Squares  and  equilateral  triangles  ? 

Ex.  19.  Squares  and  regular  pentagons  ? 

Ex.  20.  Squares  and  regular  hexagons  ? 

Ex.  21.  Squares  and  regular  heptagons  ? 

Ex.  22.  Regular  pentagons  ? 

Ex.  23.  Regular  pentagons  and  equilateral  triangles  ? 

Ex.  24.  Regular  pentagons  and  regular  hexagons  ? 

Ex.  25.  Regular  hexagons  and  equilateral  triangles  ? 

Ex.  26.  Regular  heptagons  and  equilateral  triangles  ? 

Ex.  27.  Show  that  if  two  trihedrals  have  two  face  angles  and  the  in- 
cluded dihedral  of  the  first  equal  to  two  face  angles  and  the  included  dihe- 
dral of  the  second,  the  trihedrals  will  be  congruent. 


Proof.  If  ZASB=ZETF,ZASC=ZETG,  a.i\dB-AS-C=F-ET-G, 
place  trihedral  T  on  trihedral  S  so  that  Z  ETG  shall  coincide  with  Z  ASC, 
TE  falling  along  8 A. 

Plane  ETF  falls  in  plane  ASB.  (Why  ?) 

TF  falls  along  SB.  (Why  ?) 

(See  proof  of  V.  1.) 


XXII.     PLANAL  ANGLES 


265 


Ex.  28.    Show  by  superposition  that :  . 

If  two  trihedrals  have  two  dihedrals  and  the  included  face  angle  of  the 
first  equal  to  two  dihedrals  and  the  included  face  angle  of  the  second,  the 
trihedrals  will  be  congruent.  s 

(See  proof  of  V.  2.) 

Ex.  29.    An  isosceles  trihedral  is  isoangular. 

Proof.  In  the  figure,  if  Z  ASB  =  A  BSC,  then 
B-SA-C  =  B-8C-A.  A 

Bisect  A-SB-C  by  the  plane  BSE.  Then  use  lOx.  27 
to  show  the  trihedrals  S-ABE  and  S-GBE  congruent. 


XXIII.     GROUP   ON   THE   PRISM   AND   THE 
CYLINDER 


DEFINITIONS 
(a)  The  Prism 

A  Polyhedron  is  a  solid  bounded  by  polygons  called  Faces. 

The  Edges  of  a  polyhedron  are  the  sides  of  its.  faces. 

The  Vertices  of  a  polyhedron  are  the  vertices  of  its  faces. 

A  Section  of  a  polyhedron  is  a  polygon  obtained  by  passing  a 
plane  through  the  polyhedron. 

A  Convex  polyhedron  is  one  of  which  the  sections  are  all 
convex. 

In  all  subsequent  definitions  polyhedra  will  be  assumed  to 
be  convex. 

A  Regular  polyhedron  is  one  whose  faces  are  congruent 
regular  polygons  and  whose  polyhedrals  are  congruent. 

Polyhedra  are  classified  according  to  the  number  of  fojces. 

A  Tetrahedron  is  a  polyhedron  of  four  faces. 

A  Hexahedron  is  a  polyhedron  of  six  faces. 

An  Octahedron  is  a  polyhedron  of  eight  faces. 

A  Dodecahedron  is  a  polyhedron  of  twelve  faces. 

An  Icosahedron  is  a  polyhedron  of  twenty  faces. 


ICOSABKORON  DODECAHEDRON  OCTAHEDRON  HEXAHEDRON         TETRAHEDRON 

A  Prism  is  a  polyhedron,  two  of  whose  faces  (called  bases)  are 
parallel  polygons,  and  whose  lateral  faces  are  parallelograms 
whose  vertices  are  all  vertices  of  the  respective  bases. 

266 


XXIII.     THE   PRISM  AND   THE   CYLINDER  267 

The  Lateral  Area  of  a  prism  is  the  sum  of  the  areas  of  the 
lateral  faces. 

The  Lateral  Edges  of  the  prism  are  the  edges  in  which  the 
lateral  faces  intersect. 

The  Altitude  of  a  prism  is  the  perpendicular  distance  between 
the  bases. 

Prisms  are  classified  according  to  the  number  of  sides  of  the 
bases. 

A  Triangular  prism  is  a  prism  whose  base  is  a  triangle. 

A  Quadrangular  prism  is  a  prism  whose  base  is  a  quadrilateral. 

A  Pentagonal  prism  is  a  prism  whose  base  is  a  pentagon. 

A  Right  Section  of  a  prism  is  a  section  made  by  a  plane  per- 
pendicular to  a  lateral  edge. 

A  Right  prism  is  a  prism  in  which  the  lateral  edges  are  per- 
pendicular to  the  bases. 

An  Oblique  prism  is  a  prism  whose  lateral  edges  are  oblique 
to  the  bases. 

A  Regular  prism  is  a  right  prism  in  which  the  bases  are 
regular  polygons. 

A  Parallelepiped  is  a  prism  in  which  the  bases  are  parallelo- 
grams ;  that  is,  a  prism  all  of  whose  faces  are  parallelograms. 

A  Rectangular  parallelepiped  is  a  right  parallelepiped  in 
which  the  bases  are  rectangles ;  that  is,  one  in  which  all  the 
faces  are  rectangles. 

A  Cube  is  a  regular  parallelepiped  in  which  the  lateral  faces 
are  squares. 

Corollaries  of  the  Definitions 

(a)  The  lateral  edges  of  a  prism  are  equal  and  parallel. 
(6)  The  faces  of  a  cube  are  equal  squares. 

A  Truncated  prism  is  that  portion  of  a 
prism  which  is  comprised  between  either 
base  and  a  section  not  parallel  to  it. 

A  Right  Truncated  prism  is  a  portion  of 
a  right  prism  comprised  between  either 
base  and  a  section  not  parallel  to  it. 


268  THE   ELEMENTS  OF  GEOMETRY 

The  Volume  of  a  polyhedron  is  its  ratio  to  some  other  poly 
hedron,  called  the  unit  of  volume. 

The  Unit  of  Volume  usually  taken  is  the  cube,  each  edge  of 
which  equals  the  unit  of  length. 

Two  polyhedra  are  said  to  be  equal  when  their  volumes  are 
equal. 

Two  polyhedra  are  said  to  be  congruent  when  they  may  be 
placed  in  coincident  superposition. 

Similar  polyhedra  are  polyhedra  that  have  the  same  number 
of  faces,  which  are  similar,  each  to  each,  and  similarly  placed. 

PROPOSITIONS 
(a)  The  Prism 
XXIII.  1.  Parallel  sections  of  a  prism  are  congruent. 


Hyp.     If,  in  the  prism  AB,  section  CEFQM  is  parallel  to 
section  HIJKL, 

Cone. :  then  CEF  •  •  •  ^  HTJ  •  •  -. 

Dem.  HIW  CE,  IJW  EF,  etc.  (Def.  of  II  pis.) 

.-.  Z  HIJ=  Z  CEF,  Z  IJK=  EFG,  etc.     (XXI.  7.) 
Now  HI=  CE,  IJ=  EF,  etc.  (VI.  1  a.) 


XXIII.     THE  PRISM  AND  THE   CYLINDER  269 

.'.  as  the  two  sections  are  mutually  equiangular  and  equi- 
lateral, they  may  be  placed  in  coincident  superposition  and  are 
therefore  congruent.  (Del  of  ^  figs.) 

Q.E.D. 

XXIII.  1  a.    The  bases  of  a  prism  are  co7igruent. 

Ex.  1.  Find  the  volume  of  a  rectangular  parallelepiped  20  ft.  long,  3  ft. 
wide,  and  5  ft.  high. 

Ex.  2.  A  bushel  contains  2150.4  cu.  in.  Find  the  height  of  a  bin  the 
bottom  of  which  is  25  ft.  by  15  ft.,  and  the  capacity  of  which  is  2500  bu. 

Ex.  3.  A  "lumber  foot"  is  12  in.  square  and  1  in.  thick.  At  $18  per 
M.  how  much  will  it  cost  to  build  a  cubical  bin  of  three-inch  lumber  to 
contain  500  bushels,  allowing  |  extra  material  for  studding  and  $10  for 
labor  ? 

Ex.  4.  The  number  of  cubic  feet  in  the  volume  of  a  cubical  block  is 
equal  to  the  number  of  square  feet  in  its  entire  surface.  Find  the  length 
of  the  edge  of  the  block. 

Ex.  5.  The  edges  of  a  rectangular  parallelepiped  are  6  ft.,  10  ft.,  and 
15  ft.  ;  the  edges  of  a  second  are  12  ft.,  14  ft.,  and  18  ft.  Find  the  edge 
of  a  cube  whose  volume  equals  the  sum  of  the  volumes  of  the  parallele- 
pipeds. 

Ex.  6.    A  rectangular  parallelepiped  is  often  called  an  "oblong  block." 

The  dimensions  of  an  oblong  block  are  in  the  ratio  of  2:3:5.  The 
number  of  cubic  feet  in  its  volume  is  10  times  the  number  of  square  feet 
in  its  entire  surface.     Find  its  dimensions. 

Ex.  7.  The  diagonal  of  one  cube  equals  the  edge  of  a  second.  Find 
the  ratio  of  the  volumes  of  the  cubes. 

Ex.  8.  The  diagonal  of  one  cube  is  3  times  as  long  as  the  edge  of  a 
second.     What  is  the  ratio  of  the  surfaces  of  the  two  cubes  ? 

Ex.  9.  The  diagonal  of  one  cube  is  a  times  as  long  as  the  edge  of 
a  second.     What  is  the  ratio  of  the  volumes  of  the  two  figures  ? 

Ex.  10.  The  volume  of  a  prism  is  324  cu.  ft.  Its  altitude  is  36  ft. 
What  is  the  area  of  the  base  ? 

Ex.  11.  The  altitude  of  a  prism  is  20  yd.  Its  base  is  an  equilateral 
triangle  each  side  of  which  is  15  ft.     Find  the  volume  of  the  prism. 

Ex.  12.  The  altitude  of  a  regular  prism  is  10,  Its  base  is  a  hexagon 
each  side  of  which  is  6.  Find  the  total  surface  and  the  volume  of  the 
prism. 


270 


THE  ELEMENTS  OF  GEOMETRY 


XXIII.  2.  Two  right  truncated  prisms  are  congruent, 
if  three  faces  including  a  trihedral  of  one  are  congruent 
respectively  to  three  faces  including  a  trihedral  of  the 
other  and  are  similarly  placed. 


Hyp.  If  the  right  truncated  prisms  AM  and  A'3r  have  the 
three  faces  of  trihedral  B  congruent  with  the  three  faces  of 
trihedral  B'  and  the  faces  are  similarly  placed, 

Cone. :  then  right  truncated  prism  AM  is  congruent  to  right 
truncated  prism  A'3f'. 

Dem.  Place  the  base  of  AM  in  coincident  superposition  with 
the  base  of  A'M' ;  AB  falling  on  A'B'. 

Then  AD  must  fall  on  A'D\  BR  on  B'R',  etc.         (XXI.  2.) 

Then  D  must  fall  on  Z>',  R  on  R',  and  Q  on  Q'. 

(These  faxies  are  ^  by  hyp.) 

.*.  the  plane  of  D,  R,  and  Q  must  fall  on  the  plane  of  D',  R\ 
and  Q'.  (Def.  of  plane  (b).) 

And  as  the  truncated  prisms  are  right,  Jf  must  fall  on  M'  and 
L  on  L'.  (XXI.  2.) 

.-.  the  right  truncated  prism  AM  is  congruent  with  right 

truncated  prism  A'M'. 

'^  Q.E.D. 


XXIII.     THE   PRISM   AND   THE   CYLINDER  271 

XXIII.  2  a.  2\vo  right  prisms  having  congruent  bases 
and  equal  altitudes  are  congruent. 

XXIII.  3.  Any  oblique  prism  is  equal  to  a  right 
prism  of  ivhich  the  altitude  equals  a  lateral  edge  of  the 
oblique  prism  and  the  bases  are  right  sections  of  the 
oblique  prism. 


Hyp.  If  the  right  prism  GJ'  has  its  bases  right  sections  of 
the  oblique  prism  AE'  and  its  altitude  J  J'  equal  to  the  edge 
EE', 

Cone. :  then  right  prism  GJ^  equals  oblique  prism  AE', 

Dem.     yle/and  A' J'  are  right  truncated  prisms.  (Const.) 

Their  bases  A-E  and  A-E'  are  congruent.       (XXIII.  1  a.) 
The  lateral  faces  BG  and  B'G'  are  congruent. 
Likewise,  the  lateral  faces  BI  and  BT  are  congruent. 

.-.  Et.  Tr.  Prism  AJ^  Rt.  Tr.  Prism  A' J'.        (XXIII.  2.) 

To  each  of  the  right  truncated  prisms  add  the  right  truncated 
prism  GE',  and  we  have 

right  prism  GJ^  equals  oblique  prism  AE'.     (Ax.  2.) 

Q.E.D. 


272 


THE  ELEMENTS  OF  GEOMETRY 


XXIII.  4.   Two  rectangular  parallelepipeds  that  have 
equal  bases  are  to  each  other  as  their  altitudes. 


^ 


Q' 


IP 


A' 

Hyp.  If  the  two  rectangular  parallelepipeds  Q  and  Q'  have 
equal  bases  and  their  altitudes  are  AB  and  A'B', 

Cone. :  then  rectangular  parallelepiped  Q :  rectangular  paral- 
lelepiped Q'l'.AB:  A'B'. 

Case  I.     AB  and  A'B'  commensurable. 

Case  II.     AB  and  A'B'  incommensurable. 

Dem.     Case  I.     Find  a  common  measure  of  AB  and  A'B'. 
Let  it  be  contained  in  AB  five,  and  in  A'B'  three  times. 

Then  AB :  A'B' : :  5  :  3. 

Through  the  points  of  division  draw  planes  parallel  to  the 
bases. 

The  small  rectangular  parallelepipeds  thus  obtained  are  all 
congruent.  (XXIII.  2  a.) 

In  Q  there  are  five,  in  Q'  three  of  these  equal  parallelepipeds. 

.-.  Q:  Q'l'.AB:  A'B'.  (Ax.  1.) 

Q.E.D. 

Ex.  13.  The  volume  of  a  regular  octagonal  prism  of  altitude  8  is  equal 
to  the  volume  of  a  regular  hexagonal  prism  of  altitude  12.  The  radius  of 
the  base  of  the  octagonal  prism  is  6.  Find  the  lateral  surface  of  each 
prism. 


XXm.     THE   PRISM  AND   THE   CYLINDER  273 

Q 

Sir" 

Q         F 


llllii^^Hiiiiii 


Dem.     Case  II.    Divide  AB  into  any  number  of  equal  parts. 
Suppose  one  of  these  equal  parts  is  contained  in  AB^  three 
times,  with  a  remainder  MB\ 

Through  M  pass  a  plane  parallel  to  the  base. 

Then  ^^  .Q.:  AM :  AB.  (Case  I.) 

If  the  number  of  equal  parts  in  AB  be  indefinitely  increased, 
the  remainder  JfJ3'  will  be  indefinitely  decreased,  but  can  never 
equal  zero,  because  AB  and  A'B^  are  incommensurable. 

.-.  A'M  approaches  AB'  as  a  limit,  and  Q"  approaches  Q'  as 
a  limit. 

limit,  and  -^  approaches  ^  as 

a  limit. 

But  -^  is  always  equal  to  AB^ :  AB,  (Case  I.) 

.*.  the  limits  of  the  variables  being  equal, 

Et.  parallelepiped  Q' :  rt.  parallelepiped  Q-.-.AM'.  AB. 

Q.E.D. 

ScH.  Two  rectangular  parallelepipeds  which  have  two  dimen- 
sions in  common  are  to  each  other  as  their  third  dimensions. 


AM  ,       AB' 

— —  approaches  — - —  as  a 
AB     ^^  AB 


Ex.  14.  The  volume  of  each  of  two  prisms  is  1386  cu.  ft.  The  base  of 
the  first  is  an  equilateral  triangle  whose  altitude  is  20  ft.  The  base  of  the 
second  is  a  square,  each  side  of  which  is  20  ft.  Find  the  ratio  of  the 
altitudes  of  the  prisms. 


274 


THE  ELEMENTS  OF  GEOMETRY 


XXIII.   4  a.    Tloo  rectangular  parallelepipeds  which 
have  equal  altitudes  are  to  each  other  as  their  bases. 


A 

/ 

/r 

/ 

/I;/ 

/ 
/ 

f 

l>' 

A— 

/ 

^• 

/ 

/ 

/ 

^ 

Hyp.     If  the  rectangular  parallelepipeds  Q  and  Q'  have  their 
altitudes  equal,  and  the  base  of  Q,  a  •  6  and  of  Q',  c  •  e, 


Cone. :  then 


Q  :  Q' : :  a  .  &  :  c  .  e. 


Dem.     Construct  a  rectangular  parallelepiped  Q"  whose  alti- 
tude is  ^,  and  whose  base  is  a  •  e. 


Then  Q :  Q"  : :  6  :  e.  (XXIII.  4.  Sch.) 

But  Q"  :  Q' : :  a  :  c.  (XXIII.  4.  Sch.) 

.-.  Q :  Q' : :  a  .  6  :  c  .  e.  (By  mult.) 

Q.E.D. 

XXIII.    5.    Two  rectangular  parallelepipeds  are  to 
each  other  as  the  products  of  their  three  dimensions, 

Q 


Hyp.     If  Q  and  Q'  are 

two  rectangular  paral- 
lelepipeds whose  bases 
are  a  •  h  and  a'  •  h\  re- 
spectively, and  whose 
altitudes  are  c  and  c', 
respectively. 


Cone  :  then         Q :  Q' : :  a  •  6  •  c :  a'  •  6'  •  c'. 

Dem.    Construct  a  rectangular  parallelepiped  Jf,  whose  base 
is  a  •  6  and  whose  altitude  is  c'. 


XXIII.     THE   PRISM  AND   THE   CYLINDER  275 


Then  Q 

And  M 

.-.  Q 


M::c:c'.  (XXIII.  4.) 

Q'::a'b:a'  'b\  (XXIII.  4  a.) 

Q'::a'b'c:a''b'  'C',  (By  mult.) 

Q.E.D. 

XXIII.  5  a.  Tlie  volume  of  a  rectangular  parallele- 
piped equals  the  product  of  its  three  dimensions. 

-Q- 
Hyp.     If  Q  is  a  rec- 
tangular parallelepiped 
whose    dimensions    are 
a,  ft,  and  c  ■""ilMIHi 

Cone. :  then  volume  of  Q  =  a  •  6  •  c. 

Dem.     Construct  a  cube  U,  whose  edge  is  the  linear  unit. 

Then  Q:U::a-b'C:l -1-1.  (XXIII.  5.) 

But  Q:U  is  the  volume  of  Q.  (Del  of  vol.) 

AT                              a '  b  •  c  J 

And —  =  a  -b  '  c. 

.\  the  volume  of  Q  =  a  •  6  •  c. 

Q.E.D. 

ScH.  The  volume  of  a  rectangular  parallelepiped  equals  the 
product  of  its  base  by  its  altitude. 

Ex.  15.  The  interior  dimensions  of  a  water  tank  are  8  ft.,  4  ft.,  and 
5  ft.,  respectively.     How  many  gallons  will  the  tank  hold  ? 

Ex.  16.  How  much  will  it  cost  to  line  the  tank  with  zinc  at  S5^  a 
square  yard,  allowing  a  waste  of  y\  of  the  material  for  seams  ? 

Ex.  17.  The  convex  surface  of  a  right  circular  cylinder  is  equal  to  the 
total  surface  of  a  cube ;  the  diameter  of  the  cylinder  and  its  altitude  each 
equals  10.     Find  the  edge  of  the  cube. 

Ex.  18.  The  altitude  of  a  right  circular  cylinder  is  a,  the  radius  of  its 
base  is  b.  To  find  the  radius  of  a  circle  equal  in  area  to  the  convex  sur- 
face of  the  cylinder. 


276 


THE  ELEMENTS  OF  GEOMETRY 


XXIII.  6.   The  volume  of  any  parallelepiped  equals 
the  product  of  its  base  and  its  altitude. 


Hyp.  If  P'H  is  any  parallelepiped  whose  base  is  the  paral- 
lelogram P-H'  and  whose  altitude  is  the  perpendicular  between 
the  bases, 

Cone. :  then  the  volume  of  PH  equals  the  area  of  its  base 
times  its  altitude. 

Dem.  Produce  PO,  making  KL  =  PO,  and  through  K  and  L 
pass  planes  J'JIif  and  Q'Nl.  KL. 

Extend  the  faces  HP,  H'P',  PO',  and  GH'  to  intersect  the 
planes  J'3/ and  Q'N,  forming  the  right  parallelepiped  MQ'. 

Oblique  parallelepiped  P'H  =  right  parallelepiped  MQ'. 

(XXIII.  3.) 

Again,  produce  N'Q'  making  Q'F'  =  N'Q'  and  through  Q'  and 
F'  pass  planes  Q'Jand  F'R±  Q'F'. 

Extend  the  faces  Q'M'  and  LM,  Q'N  and  M'K  to  intersect 
the  planes  Q'e/and  F'R,  forming  the  rectangular  parallelepiped 
Q'R.  (Def.  of  rt.  parallelepiped.) 

Rt.  parallelepiped  MQ'=TQQ,t.  parallelepiped  Q'R.  (XXIII.  3.) 

.•.  rect.  parallelepiped  Q'J?=obl.  parallelepiped  P'H.  (Ax.  1.) 

Now  the  volume  of  a  rectangular  parallelepiped  equals  the 
product  of  the  base  and  an  altitude.  (XXIII.  5  a.) 

And  as  the  base  and  an  altitude  of  Q'R  equal,  respectively, 

the  base  and  altitude  of  PH,  the  volume  of  PH  equals  the 

area  of  its  base  times  the  altitude. 

Q.E.O. 


XXIII.     THE   PRISM  AND   THE   CYLINDER  277 

XXIII.  7.  The  plane  through  two  diagonally  opposite 
edges  of  a  parallelepiped  divides  the  figure  into  two 
equal  triangular  prisms.  yj 

Hyp.  If  AG  is  a  parallelepiped  A^'/^/^-^--^J  / 
and  a  plane  ACOR  is  passed  through  'yT^-^^lZ^^rf/f 
AR  and  CQy  /  eI _jt 


Cone.  •  then  triangular  prism  ACB-F  equals  triangular  prism 
ACE-H. 

Dam.  Through  S,  any  point  of  ATI,  pass  a  plane  SMQK 
perpendicular  to  RA  and  intersecting  ACOR  in  SQ. 

Plane  BR  is  parallel  to  plane  CH.     (Def.  of  parallelepiped.) 

.-.  SKW  MQ,  (Def.  of  II  planes.) 

Similarly,  "     SM  II  KQ. 

.*.  the  4-side  S-Q  is  a  parallelogram.     (Def.  of  O.) 

.-.  A  SKQ  ^  A  SQM.  (VI.  1  a.  Sch.) 

Prism  ABC-F  equals  a  right  prism  whose  base  is  A  SKQ 
and  whose  altitude  is  FB.  (XXIII.  3.) 

Prism  ACE-H  equals  a  right  prism  whose  base  is  A  SRQ 
and  whose  altitude  is  FB  (or  EH).  (XXIII.  3.) 

But  these  right  prisms  are  equal.  (XXIII.  2  a.) 

.'.  triangular  prism  ACB-F  equals  triangular  prism  ACE-H. 

(Ax.  1.) 
Q.E.D. 

Ex.  19.  The  total  surface  of  a  right  circular  cylinder  whose  height  is 
twice  the  radius  of  its  base  is  equal  to  the  surface  of  a  cube.  If  the  edge 
of  the  cube  is  12  in.,  what  is  the  lieight  of  the  cylinder  ? 

Ex.  20.  The  altitude  of  a  cylinder  is  12  ft.  Its  base  is  a  circle  of 
radius  8  in.     Find  the  volume  and  the  total  surface  of  the  cylinder. 

Ex.  21.  Find  the  diameter  of  a  cylindrical  tank  10  ft.  deep,  whose 
capacity  is  8000  gal. 


278         THE  ELEMENTS  OF  GEOMETRY 

XXIII.  8.   The  volume  of  a  triangular  prism  equals 
the  product  of  its  base  by  its  altitude. 


Hyp.  If  ABG-F  is  a  triangular  prism  whose  base  is  ABG 
and  whose  altitude  is  ET^ 

Cone. :  then  the  volume  of  AGB-F=A  ABG  x  ET. 

Dem.     Complete  the  parallelograms  ABGO  and  EFQH. 

Draw  OS"  completing  the  parallelepiped  ABGO-F. 

The  volume  of  parallelepiped  ABGO-F  =^n  ABGO  x  ET. 

(XXIII.  6.) 

But  the  volume  of  prism  AGB-F  equals  one  half  that  of 

parallelepiped  AGBO-F.  (XXIII.  7.) 

And  A  ABG==  \  n  ABGO.  (VI.  1  a.  Sch.) 

.-.  volume  of  prism  AGB-F  =  A  ABG  x  ET.  (Ax.  3.) 

Q.E.D. 

Ex.  22.  How  much  sheet  iron  would  it  take 'to  make  such  a  tank 
(Ex.  21),  allowing  y^  for  waste  and  seams  ? 

Ex.  23.  A  cylindrical  pipe  of  diameter  20  in.  discharges  800  gal.  a 
second.    What  is  the  velocity  of  the  water  in  the  pipe  ? 

Ex.  24,  The  total  surface  of  a  right  circular  cylinder  whose  height  is 
three  times  the  diameter  of  its  base  is  2513.28  sq.  ft.  Find  the  volume  of 
the  cylinder. 


XXIII.     THE   PRISM  AND   THE   CYLINDER  279 

XXIII.   8  a.    The  volume  of  any  prism  equals   the 
product  of  its  base  by  its  altitude. 


Hyp.  If  ABCO-F  is  any  prism,  and  ABCOJ  is  its  base,  and 
ET  is  its  altitude, 

Cone. :  then  its  volume  equals  the  product  of  its  base  times 
its  altitude. 

Dem.     Through  EA  and  OH,  EA  and  CQ,  pass  planes. 

IHiese  planes  divide  the  prism  into  triangular  prisms. 

The  volume  of  each  triangular  prism  equals  its  base  times 
its  altitude.  (XXIII.  8.) 

But  the  sum  of  the  bases  of  the  triangular  prisms  equals  the 
base  of  the  given  prism,  and  the  altitude  of  the  triangular  prisms 
is  the  altitude  of  the  given  prism. 

.*.  the  volume  of  prism  J.S(70-jP  equals  the  product  of  its 
base  times  its  altitude. 

Q.E.D. 

Ex.  25.  The  diameter  and  the  altitude  of  a  cylinder  are  each  equal  to 
the  edge  of  a  cube.     What  is  the  ratio  of  the  volumes  of  the  two  figures  ? 

Ex.  26.    A  cubic  foot  of  cast  iron  weighs  445  lb.     What  is  the  weight 

:  a  cast-iron  pipe 
nal  measurement  ? 


280  th:e  elements  of  geometry 

XXIII.  8  b.  Any  two  prisms  are  to  each  other  as  the 
products  of  the  bases  by  the  altitudes. 

If  the  bases  are  equal,  the  prisms  are  to  each  other  as 
the  altitudes. 

If  the  altitudes  are  equal,  the  prisma  are  to  each  other 
as  the  bases. 

If  the  bases  are  equal  and  also  the  altitudes,  the 
prisms  are  equal. 

XXIII.     SUMMARY  OF  PROPOSITIONS    IN   THE  GROUP 
ON   (a)   THE  PRISM 

1.  Parallel  sections  of  a  prism  are  congruent 
a.   The  bases  of  a  prism  are  congruent 

2.  Tioo  right  truncated  prisms  are  congruent,  if  three 
faces  including  a  trihedral  of  the  one  are  equal  respeo- 
tively  to  three  faces  including  a  trihedral  of  the  other 
and  are  similarly  placed. 

a.   Two  right  prisms  having  equal  bases  and  equal 
altitudes  are  congruent. 

3.  Any  oblique  prism  is  equal  to  a  right  prism  of 
ichich  the  altitude  equals  a  lateral  edge  of  the  oblique 
prism,  and  the  bases  are  right  sections  of  the  oblique 
prism. 

4-  Tioo  rectangidar  parallelepipeds  that  have  equal 
bases  are  to  each  other  as  their  altitudes. 

a.  Two  rectangidar  parallelepipeds  that  have  eqwA 
altitudes  are  to  each  other  «5  their  bases^ 


XXIII.     THE   PRISM  AND  THE   CYLINDER  281 

5.  Two  rectangular  parallelepipeds  are  to  each  other 
as  the  products  of  their  three  dimensions. 

a.   The   volume   of  a   rectangular  parallelepiped 
equals  the  product  of  its  three  dimensions. 

6.  The  volume  of  any  parallelepiped  equals  the  prod- 
uct of  its  base  by  its  altitude. 

7.  The  plane  through  two  diagonally  opposite  edges 
of  a  parallelejnped  divides  the  figure  into  ttvo  equal 
triangular  prisms. 

8.  The  volume  of  a  triangular  prism  equals  the  prod- 
uct of  its  base  by  its  altitude. 

a.  The  volume  of  any  prism  equals  the  product  of 

its  base  by  its  altitude. 

b.  Any  two  prisms  are  to  each  other  as  the  products 

of  the  bases  by  the  altitudes. 
If  the  bases  are  equals  the  prisms  are  to  each 

other  as  the  altitudes. 
If  the  altitudes  are  equals  the  prisms  are  to 

each  other  as  the  bases. 
Jf  the  bases  are  equal,  and  also  the  altitudes, 

the  prisms  are  equal. 


282  THE   ELEMENTS  OF  GEOMETRY 

DEFINITIONS 
(b)  The  Cylinder 

A  Cylindrical  Surface  is  a  surface  generated  by  a  straight  line 
that  moves  parallel  to  its  first  position  along  a  curve  not  co- 
planar  with  the  moving  line. 

The  moving  line  is  called  the  Generatrix. 

The  curve  that  directs  the  motion  is  called  the  Directrix. 

The  successive  positions  of  the  generatrix  are  called  the 
Elements  of  the  Surface. 

A  Cylinder  is  a  solid  inclosed  by  a  cylindrical  surface  and 
two  parallel  planes. 

The  Bases  of  a  cylinder  are  the  parallel  plane  sections. 

The  Elements  of  the  Cylinder  are  the  portions  of  the  elements 
of  the  cylindrical  surface  determined  by  the  bases. 

The  Altitude  of  a  cylinder  is  the  perpendicular  distance 
between  the  bases. 

A  Right  Section  of  a  cylinder  is  a  section  made  by  a  plane 
perpendicular  to  an  element. 


Cylindbbs 

A  Right  cylinder  is  a  cylinder  the  elements  of  which  are 
perpendicular  to  the  bases. 

A  Circular  cylinder  is  a  cylinder  the  bases  of  which  are 
circles. 

A  point  is  said  to  revolve  around  a  fixed  line  when  it  gen- 
erates a  circle  whose  plane  is  perpendicular  to  the  fixed  line 
and  having  its  center  on  the  fixed  line. 


XXIII.     THE   PRISM   AND   THE   CYLINDER  283 

The  fixed  line  is  called  the  Axis  of  Revolution,  or  simply  the 
axis. 

A  line  or  surface  is  said  to  revolve  about  the  axis,  when 
every  point  in  the  moving  line  or  surface  revolves  about  the 
axis. 

The  surface  generated  by  the  revolution  of  a  line  (straight 
or  curved)  about  an  axis  is  called  a  Surface  of  Revolution. 

The  volume  (or  solid)  generated  by  the  revolution  of  a  sur- 
face about  an  axis  is  called  a  Volume  (or  Solid)  of  Revolution. 

The  axis  of  revolution  is  often  called  the  Axis  of  the  Surface 
or  Volume  generated  by  the  revolution. 

Corollaries  of  the  Definitions 

(a)  Any  section  of  a  cylinder  through  an  eleinent  is  a 
parallelogrmiv. 
(h)  A  right  circular  cylinder  is  a  cylinder  of  revolution. 

As  the  circle  has  been  shown  (XVIII)  to  be  the  limit  of  the 
regular  polygon  as  the  number  of  sides  is  increased  indefi- 
nitely, so  the  circular  cylinder  is  the  limit  in  surface  and  vol- 
ume of  the  prism  with  regular  bases,  as  the  number  of  sides  of 
the  bases  is  increased  beyond  any  assignable  number.  (It  will 
be  a  good  exercise  for  the  student  to  give  the  detailed  proof  of 
the  statement.) 

Accordingly,  every  proposition  that  is  true  of  every  prism 
with  a  regular  base,  whatever  may  he  the  number  of  lateral  faces, 
is  true  of  the  circular  cylinder.  We  therefore  obtain  from  the 
corresponding  propositions  of  XXIII  (a),  the  summary  on  the 
following  page. 

Ex.  27.  A  horse  power  is  the  force  necessary  to  raise  33,000  lb.  1  ft.  in 
1  min.  The  cylinder  of  an  engine  is  4  ft.  in  diameter  and  6  ft.  high  ;  the 
piston  is  6  in.  thick  and  the  piston-rod  8  in.  in  diameter. 

Find  the  horse  power  of  the  engine  when  it  is  making  200  revolutions  a 
minute  with  a  steam  pressure  of  60  lb.  to  the  square  inch. 

Ex.  28.  Show  that  no  polyhedron  can  have  less  than  four  faces  nor  lesB 
than  six  edges. 


284  THE  ELEMENTS  OF  GEOMETRY 


ZXni.     SUMMART   OF    PROPOSITIONS   IN   THE   GROUP 
ON   (6)    THE  CIRCULAR   CYLINDER 

1.  Parallel  sections  of  a  circular  cylinder  are  con- 
gruent 

a.   The  bases  of  a  circular  cylinder  are  congruent. 

2.  The  volume  of  a  circular  cylinder  equals  the  prod- 
uct of  the  area  of  its  base  by  its  altitude. 

a.  Jf  B.  be  the  altitude  of  any  circular  cylinder, 

and  R  the  radius  of  either  base,  the  volume 
of  the  cylinder  equals  irR  H. 

b.  If  B.  be  the  altitude  of  any  rigid  circular  cylin- 

der, and  R  the  radius  of  either  base,  the  area 
of  the  convex  surface  of  the  cylinder  equals 
2irRH. 

c.  The  volumes  of  any  tioo  circular  cylinders  are 

to  each  other  as  the  products  of  the  areas  of 

the  bases  by  the  altitudes. 
If  the  bases  are  equal,  the  cylinders  are  to  each 

other  as  the  altitudes. 
If  the  altitudes  are  equal,  the  cylinders  are  to 

each  other  as  the  areas  of  the  bases. 
If  the  bases  are  equal  and  also  the  altitudes, 

the  cylinders  are  equal. 


XXin.    THE   PRISM  AND   THE   CYLINDER  285 

Ex.  29.  Every  plane  section  of  a  parallelepiped  is  a  parallelogram  if 
the  plane  of  tiie  secLion  intersects  4  parallel  edges. 

Ex.  30.  What  kind  of  quadrilateral  is  cut  from  a  parallelepiped  by  a 
diagonal  plane  ?     How  do  the  diagonals  of  the  section  cut  each  other  ? 

Ex.  31.  Show  that  the  diagonals  of  a  parallelepiped  concur  in  a  point 
at  which  each  is  bi.sected. 

Def.  The  point  in  which  the  diagonals  of  a  parallelepiped  concur  is 
called  the  center  of  the  parallelepiped. 

Ex,  32,   Show  that  the  converse  of  this  proposition  is  true. 

Ex,  33.  Show  that  any  line  that  passes  through  the  center  {K)  of  a 
parallelepiped  and  terminates  in  opposite  faces  of  the  parallelepiped  is 
bisected  at  K. 

Ex,  34,  Show  that  the  sum  of  the  squares  of  the  diagonals  of  a  paral- 
lelepiped equals  the  sum  of  the  squares  of  the  edges. 

Ex.  35,  Show  that  the  diagonals  of  a  rectangular  parallelepiped  are 
equal. 

Ex,  36.    Prove  that  the  converse  of  the  proposition  is  also  true. 

Ex,  37.  The  edges  of  a  rectangular  parallelepiped  are  12,  15,  and  20  ft., 
respectively.     What  is  the  length  of  the  diagonal  ? 

Ex.  38.  Show  how  to  construct  a  parallelepiped  that  shall  have  its 
edges  on  three  given  straight  lines. 

Ex,  39.  A  plane  through  any  edge  of  the  upper  base  and  the  diagonally 
opposite  edge  of  the  lower  base  of  a  prism  cuts  from  the  figure  a  rectangle. 
What  kind  of  prism  is  the  original  figure  ? 

Ex.  40,  Every  plane,  through  an  edge  of  the  upper  base,  that  cuts  the 
lower  base  of  a  prism,  cuts  from  the  prism  a  parallelogram.  What  kind 
of  prism  is  the  original  figure  ? 

Ex,  41,  The  volume  of  any  regular  prism  is  equal  to  the  product  of  the 
lateral  area  by  the  apothem  of  either  base. 

Ex.  42.    Show  how  to  cut  from  a  cube  a  regular  hexagon. 

Ex.  43.  Show  that  if  a  prism  be  cut  by  two  ^  planes  and  the  cor- 
responding sides  of  the  sections  be  produced,  the  points  of  intersection  of 
these  sides  will  be  collinear. 

Ex.  44.  Show  that  the  volume  of  a  prism  equals  the  product  of  the 
area  of  a  right  section  by  the  length  of  a  lateral  edge. 

Ex.  45.  The  total  surface  of  a  circular  cylinder  is  equal  to  the  convex 
surface  of  a  cylinder  having  the  same  base  as  the  given  cylinder,  and 
having  an  altitude  equal  to  the  altitude  of  the  given  cylinder  plus  the 
radius  of  the  base. 


286         THE  ELEMENTS  OF  GEOMETRY 

Ex.  46.   The  volume  of  a  cylinder  equals  the  product  of  the  area  of  a 
right  section  by  an  element  of  the  cylinder. 

Ex.  47.    Show  that  the  volume  of  a  cylinder  is  equal  to  its  convex  sur- 
face multiplied  by  ^  the  radius  of  its  base. 

Ex.  48.   What  is  the  locus  of  a  point  whose  distance  from  a  given 
straight  line  is  equal  to  10  in.  ? 

Ex.  49.    What  is  the  locus  of  a  point  whose  distance  from  a  given  plane 
is  a  and  whose  distance  from  a  given  line  parallel  to  the  plane  is  6  ? 

Ex.  50.    What  is  the  locus  of  a  point  in  a  plane  at  a  distance  d  from  a 
line  that  intersects  the  plane  ? 

Ex.  51.    What  surface  is  generated  by  the  axis  of  a  circular  cylinder  of 
radius  a  that  rolls  on  the  inner  surface  of  a  circular  cylinder  of  radius  b  ? 

Ex.  52.    Find  a  point  equidistant  from  two  given  points,  A  and  B,  and 
also  at  a  distance  d  from  a  given  straight  line. 

Ex.  53.    Find  a  point  equidistant  from  three  given  points,  A,  B,  and  C, 
and  also  at  a  given  distance  from  a  given  straight  line. 


XXIV.  GROUP  ON  THE  PYRAMID  AND  THE 

CONE 

DEFINITIONS 

(a)   The  Pyramid 

A  Pyramid  is  a  polyhedron,  one  face  of  which  is  a  polygon, 
while  the  other  faces  are  triangles  that  have  a  common  vertex. 
This  common  vertex  is  called  the  Vertex  of  the  pyramid. 


The  Base  of  a  pyramid  is  the  polygon  on  which  the  pyramid 
is  supposed  to  rest. 

If  all  the  faces  of  a  pyramid  are  triangles,  any  one  may  be 
taken  as  the  base. 

The  faces  of  a  pyramid  other  than  the  base  are  called  the 
Lateral  Faces. 

The  sum  of  the  areas  of  the  lateral  faces  is  called  the  Lateral 
(or  Convex)  Surface  of  the  pyramid. 

The  Lateral  Edges  are  the  edges  that  meet  in  the  vertex. 

The  Altitude  of  a  pyramid  is  the  perpendicular  dropped  from 
the  vertex  to  the  base. 

287 


288         THE  ELEMENTS  OF  GEOMETRY 

Pyramids  are  said  to  be  Triangular,  Quadrangular,  Pentagonal, 
etc.,  according  to  the  number  of  sides  of  the  bases. 

A  Regular  pyramid  is  one  whose  base  is  a  regular  polygon 
that  has  for  its  center  the  foot  of  the  altitude  of  the  pyramid. 


Corollary  op  the  Definition 

(a)   The  lateral  faces  of  a  regular  pyramid  are  congruent 
isosceles  triangles. 

The  Slant  Height  of  a  regular  pyramid  is  the  altitude  of  any 
of  its  lateral  faces. 

A  Truncated  pyramid  is  that  portion  of  a        /^""^^ 
pyramid  which  is  comprised  between  the  base       /       !        \ 
and  a  plane  section  not  parallel  to  the  base.  /  ^^--^-.^    \ 


A  Frustum  of  a  pyramid  is  that  portion  of  a 
pyramid  which  is  comprised  between  the  base 
and  a  plane  section  parallel  to  the  base. 


A  Prismoid  is  a  polyhedron,  two  of  whose  fares,  called  Bases, 
are  parallel,  while  the  other  faces  (lateral  faces)  are  triangles 
or  trapezoids  that  have  their  vertices  at  the  vertices  of  the 


The  Altitude  of  a  frustum  or  of  a  prismoid  is  the  perpendicu- 
lar distance  between  the  bases. 

The  Slant  Height  of  a  reornlar  frustum  (i.e.  a  frustum  cut 
from  a  regular  pyramid)  is  the  altitude  of  any  of  its  lateral 
faces. 


XXIV.     THE  PYRAMID   AND  THE   CONE  289 

PROPOSITIONS 

(a)  Pyramids 

XXIV.  1.  If  a  set  of  lines  he  cut  hy  three  parallel 
'planes,  the  lines  are  cut  proportionally. 


Hyp.     liAB  and  CF  are  cnt  by  the  parallel  planes  SR,  FO, 
and  QM'in  B,  E,  A,  and  F,  H,  C, 

Cone. :  then  AE  :  EB::  CH:  HF. 

Dem.     Draw  the  ioin  ^F  cutting  plane  PO  in  G. 
Draw  the  joins  BF,  EG,  GH,  and  AC. 

Then 


But 


Note.  —  So  also,  AB:AE::CF:  CH, 

and  AB:BE::CF:  HF. 

ScH.     The  same  course  of  reasoning]:  may  be  extended  to  any 
number  of  lines  and  any  number  of  planes* 


AC  II  GH  and  EG  II  BF. 

(Def. 

of  11  planes.) 

\  AE.EB.'.AG:  GF. 

(XV.  1.) 

AG:  GF::  CH:  HF. 

(XV.  1.) 

•.  AE:EB::CH:HF. 

(Ax.  1.) 
Q.E.D. 

290 


THE  ELEMENTS  OF  GEOMETRY 


XXIV.  2.  Any  section  of  a  pyramid  parallel  to  the 

base  is  similar  to  the  base. 

a 


Hyp.     If,  in  the  pyramid  S-ABCER,  the  section  FOHKM  is 
piirallel  to  ABCER, 

Cone. :  then     ?i-gon  ABC n-gon  FGH  •". 

Dem.  AB  II  FG  and  RA  II  MF.     (Def.  of  II  planes.) 

.-.  Z  RAB  =  Z3IFG.  (XXI.  7.) 

Similarly,  Z  ABC  =  Z  FGH;  Z  BCE  =Z  GHK,  etc.  j 
i.e.  ABCE  and  FGHKare  mutually  equiangular. 
Again,  A  SIIG  -^ASCB; 

A  SGF  ~  A  SB  A,  etc.  ;;XV.  2.) 


And 


•.  BC'.GH 

AB.FG 

:  AB : FG 


:  ^>S' :  GS.  (Hom.  sides  of  ~  A.) 
:  jB/S'  :  (rxS'.  (Same  reason.) 

:  BC'.GH.         .  (Ax.  1.) 


Similarly,  the  other  pairs  of  homologous  sides  are  propor- 
tional 

.-.  w-gon  ABCE w-gon  FGHK -  •  -. 

(Def.  of  '^  figs.) 

Q.E.D. 

Ex.  1.  From  a  point  A  without  a  plane  obliques  are  drawn  terminating 
on  the  plane.  On  each  oblique  a  point  P  is  so  taken  as  to  divide  the  line 
in  the  ratio  of  2  :  3.     Show  that  the  locus  of  P  is  a  plane.        (XXIV.  1.) 


XXIV.     THE  PYRAMID   AND  THE   CONE 


291 


XXIV.  2  a.  The  perimeters  of  parallel  sections  of  a 
pyramid  are  to  each  other  as  the  distances  of  the  sec- 
tions from  the  vertex. 


/   I 

J 

/  ^/vM,'^ 

i 

\/   1/ 

Hyp.  If  8M\^  perpendicular  to  plane  FQ  in  3f  and  perpen- 
dicular to  plane  AE  in  i,  and  if  section  FORK-"  is  parallel 
to  section  ABCE  •  •  •, 

Cone. :  then  perim.  FQHK:  perim.  ABCE"-: :  SM:  SL. 

Dem.  Through  S  pass  a  plane  JT  parallel  to  plane  VQ 
parallel  to  plane  AE. 

Perim.  FGHK:  perim.  ABCE  -..FG:  AB.     (XXIV.  2.) 
But  FG:  AB  ::SG:  SB.     (Hom.  sides  of  ~  A.) 

And       SG:SB::SM:SL'.:SF:  SA,  etc.  (XXIV.  1.) 

.-.  perim.  FGHK:  perim.  ABCE  ::SM:  SL.  (Ax.  1.) 

Q.E.D. 


Ex.  2.  AB  and  CE  are  two  lines 
not  in  the  same  plane.  Any  number 
of  lines  QB  terminating  in  AB  and 
CE  are  bisected  at  D.  Show  that  the 
locus  of  Z)  is  a  plane  through  the  mid- 
point M  of  the  common  ±  FG. 

Pass  a  plane  through  M  parallel  to 
AB  and  CE.  Through  AB  and  CE 
pass  planes  parallel  to  the  first  plane. 
Use  XXIV.  1. 


292         THE  ELEMENTS  OF  GEOMETRY 

XXIV.  2  b.  The  areas  of  two  parallel  sections  of  a 
pyramid  are  to  each  other  as  the  squares  of  their  dis- 
tances from  the  vertex. 


7 


Hyp.  If  SM  is  perpendicular  to  plane  FQ  in  M&nd  perpen- 
dicular to  plane  AE  in  L,  and  if  section  FGHK  •••is  parallel 
to  section  ABC  •••, 

Cone :  then  area  of 

w-gon  FORK:  area  of  w-gon  ABCE : :  SM^ :  SI?. 

Dem.      w-gon  FGHK :  rj-gon  ABCE::  F(f  :  AIT.      (XVI.  3.) 

But  FG:AB::SG:SB::SM:SL.         (XXIV.  1.) 

.-.  FW  :  A&  ::SG^:SB'::  SM""  :  SL\      (XI.  (C).) 

.-.  area  of  n-gon  FGHK:  area  of  n-gon  ABCE  : :  SM^ :  SL\ 

Q.E.D. 

Ex.  3.  The  base  of  a  pyramid  is  a  regular  octagon  whose  radius  is  10  ft. 
The  altitude  of  the  pyramid  is  24  ft.  Find  tlie  perimeter  and  the  area  of 
a  section  parallel  to  the  base  and  6  ft.  from  the  vertex. 

Ex.  4.  Two  pyramids  have  the  same  volume.  The  area  of  the  base  of 
the  first  is  120  sq.  yd.  and  its  altitude  60  ft.  The  altitude  of  the  second  is 
35  ft.     What  is  the  area  of  its  base  ? 

Ex.  5.  The  altitude  of  a  regular  pyramid  is  15  ft.  Its  base  is  a  square, 
each  side  of  which  is  4  ft.     Find  the  volume  of  the  pyramid. 

Find  also  the  lateral  surface  and  the  total  surface  of  the  pyramid. 


XXIV.     THE  PYRAMID   AND  THE   CONE 


293 


XXIV.  '^  c.  If  tivo  pyramids  have  equal  altitudes  and 
equal  bases,  sections  parallel  to  their  bases  and  equally 
distant  from  their  vertices  are  equal. 

V 


Hyp.  If  A  ACE  =  A  FOH,  altitude  SL  =  altitude  VM,  and 
SV  =  VM', 

Cone:  then      n-gon  A'C'E' =  n-gon  F'G'H'. 

Dem.  Area  ACE :  area  A'C'E'  ::STJ:  SL'\  (XXIV.  2  6.) 

Area  FGH :  area  F'G'H' : :  VM'  :  VM^'. 

But  ST  :  SIJ' : :  VM' :  VM^.  (Hyp.) 

.-.  area  ACE  :  area  FGH : :  area  A'C'E' :  area  FG'H'. 

But  area  ACE  =  area  FGH. 

.«.  area  A'C'E'  =  area  i^'G^'^. 

Q.E.D. 

Note.  —  From  the  fourth  proportion  of  the  demonstration,  it  follows 
that  if  two  pyramids  have  equal  altitudes,  the  areas  of  sections  parallel  to 
the  bases  and  equidistant  from  them  have  the  same  ratio  as  the  bases. 

Ex.  6.  The  base  of  a  regular  pyramid  is  a  dodecagon  whose  radius  is 
20  in.    The  volume  of  the  pyramid  is  4000  cu.  in.     Find  the  altitude. 

Ex.  7.  Find  the  total  surface  and  the  volume  of  a  regular  triangular 
pyramid,  each  edge  of  wliich  is  a. 

Ex.  8.  Each  lateral  edge  of  a  regular  triangular  pyramid  is  29  ft. ;  the 
altitude  is  21  ft.     Find  the  total  surface  and  the  volume  of  the  pyramid. 


294 


THE  ELEMENTS  OF  GEOMETRY 


XXIV.  3.   Two  triangular  pyramids  that  have  equal 
bases  and  equal  altitudes  are  equal. 


Hyp.     If,  in  the  triangular  pyramids  E  and  E',  base  ABC  = 
base  A'B'C'j  and  if  the  altitudes  of  the  two  pyramids  are  equal, 


Cone. :  then 


Pyr.  E  =  Pyr.  E' 


Dem.     If  ^  :,fc  E',  let  E-E'  =  r. 

Divide  altitude  AT  into  any  number,  say  four,  equal  parts. 

Through  each  point  of  division  pass  parallel  sections. 

On  the  base  ABC  construct  a  prism  whose  lateral  edges  are 
parallel  to  AE  and  whose  altitude  equals  \  AT. 

Similarly,  construct  a  prism  upon  each  section  as  a  base. 

This  set  of  prisms  is  circumscribed  about  Pyr.  E. 

With  the  topmost  section  of  Pyr.  E*  as  an  upper  base,  con- 
struct a  prism  whose  lateral  edges  are  parallel  to  A'E'  and 
whose  altitude  equals  \  AT. 

Similarly,  construct  prisms  with  the  remaining  sections  as 
upper  bases. 

This  set  of  prisms  is  inscribed  in  Pyr.  E'. 

Each  prism  in  Pyr.  E'  equals  the  prism  next  above  it  in 
Pyr.  E.  (XXIII.  8  b.) 

.'.  the  difference  between  the  sums  of  prisms  of  Pyr.  E  and 
Pyr.  E'  is  the  lowest  prism  of  Pyr.  E. 


XXIV.     THE   PYRAMID   AND   THE    CONE 


295 


Let  sum  of  prisms  in  Pyr.  E  =  S,  and  sum  of  prisms  in 
Pyr.  E'  =  S',  and  let  volume  of  lowest  prism  in  E  =  v. 

Then  S-S'  =  v. 

But  E<S  and  S'  <  E\ 

.:  E  +  S'<S  +  E'.     (Preliminary  Th.  1.) 
.-.  E-E'<S-  S'.     (Preliminary  Th.  3.) 

That  is,  E~E'  <  v,  or  r  <  v.  (1) 

Now,  if  the  number  of  equal  parts  into  which  altitude  AT 
is  divided  is  increased,  the  difference,  v,  is  correspondingly 
decreased. 

Evidently,  by  increasing  the  number  of  equal  parts  of  ^T 
indefinitely,  this  difference,  or  v,  can  be  made  smaller  than 
any  assignable  value  except  zero. 

In  other  words,  v  may  be  less  than  r,  a  constant.  (2) 

But  we  have  just  proved  that  r  <v.  (1) 

.*.  unless  r  equals  zero,  (2)  contradicts  (1). 
,  .-.  r  must  equal  zero. 

That  is,  Pyr.  E  =  Pyr.  E'. 


Ex.  9.  The  edges  of 
an  oblong  block  are  a,  b, 
and  c.  The  centers  of 
the  faces  are  joined  as 
shown  in  the  figure. 
Find  the  volume  of  the 
octahedron  thus  formed. 

Ex.  10.  Join  the  cen- 
ter of  a  cube  to  its  ver- 
tices .  By  considering  the 
pyramids  thus  formed, 
show  that  the  volume  of 
a  cube  equals  one  sixth 
the  product  of  its  total 
surface  by  one  edge. 


296 


THE  ELEMENTS  OF  GEOMETRY 


XXIV.  4.   The  volume  of  a  triangular  pyramid  is 
one  third  the  product  of  its  base  and  altitude. 


Hyp.     If  F-ABC  is  a  triangular  pyramid  of  base  b  and 
altitude  h, 

Cone. :  then  volume  of  pyramid  F-ABC=^b  -  h. 

Dem.     Complete  the  triangular  prism  ABC-GFE. 

Draw  EC. 

This  prism  minus  pyramid  i^'-^SC  equals  pyramid  F-AOQE. 

Pyr.  F-ACGE  =  Pyr.  F-AEO  +  Pyr.  F-GEC. 

AAEG^AGEC, 

and  the  altitudes  of  these  two  pyramids  are  equal. 

.-.  Pyr.  F-AEG=  Pyr.  F^GEC.  (XXIV.  3.) 

But  pyramid  F-ABG  may  be  read  C-ABF,  and  pyramid 
F-AEG  may  be  read  C-AEF. 

But  Pyr.  C-ABF  ^  Pyr.  G-AEF.  (XXIV.  3.) 

.-.  Pyr.  C-ABF==  Pyr.  C-AEF  =  Pyr.  F-GEC,     (Ax.  1.) 

But  the  sum  of  these  three  pyramids  equals  the  prism. 

.*.  volume  of  prism  =  b  *h. 

.-.  volume  of  Pyr.  F-ABC=ih  «  h. 

Q.E.D. 


XXIV.     THE   PYRAMID   AND  THE   CONE  297 

XXIV.  4  a.   The  volume  of  any  pyramid  equals  one 
third  the  product  of  its  base  and  altitude. 


Note,  — Let  the  student  prove  this  corol- 
lary by  dividing  the  given  pyramid  into  tri- 
angular pyramids  with  a  common  altitude  and 
taking  their  sum. 


XXIV.  4  b.  The  volumes  of  any  two  pyramids  are  to 
each  other  as  the  products  of  their  bases  and  altitudes, 

(Let  the  student  give  the  proof,  using  XXIV.  4  a.) 

ScH.  1.  If  the  bases  are  equal,  the  pyramids  are  to  each 
other  as  their  altitudes. 

If  the  altitudes  are  equal,  the  pyramids  are  to  each  other  as 
their  bases. 

ScH.  2.  The  volume  of  any  polyhedron  may  be  found  by 
dividing  the  figure  into  pyramids  and  adding  the  volumes  of 
these  pyramids. 

Ex.  11.  The  great  pyramid  of  Cheops  is  486  ft.  high  and  its  base  is  a 
square  768  ft.  on  a  side.     Find  the  lateral  surface  in  square  yards. 

Ex,  12.  Find  the  volume  and  the  total  surface  of  a  regular  quadrangu- 
lar pyramid  each  of  whose  base  edges  is  10  ft.  and  each  of  whose  lateral 
edges  is  20  ft. 

Ex.  13.  Each  edge  of  the  base  of  a  regular  octagonal  pyramid  is  12  ft. 
Each  lateral  edge  is  40  ft.  Find  the  convex  surface  and  the  volume  of  the 
pyramid. 

Ex.  14.  The  lateral  edges  of  a  regular  pentagonal  pyramid  are  each 
5  a.  Each  side  of  the  base  is  a.  Find  the  volume  and  the  total  surface 
of  the  pyramid. 

Ex.  15.  The  altitude  of  a  square  pyramid,  each  edge  of  whose  base  is 
a,  is  equal  to  the  diagonal  of  the  base.  Find  the  volume  and  the  total 
surface  of  the  pyramid. 


298 


THE  ELEMENTS  OF  GEOMETRY 


XXIV.  5.  The  volume  of  any  triangular  frustum 
equals  one  third  the  product  of  the  altitude  into  the 
sum  of  the  two  bases  and  a  mean  proportional  betioeen 
them. 


Hyp.     If  ABC-G  is  a  triangular  frustum  of  bases  b  and  b\ 
and  of  altitude  h, 

gone. :  then  volume  of  ABC-G  =  J  A6  +  ^  /i6'  +  J  hVW 

=  ^h(b  +  b'  +  ^bb'). 

Dem.     Draw  BE,  BG,  and  EO. 

Ft.  ABC-G  =  Pyr.  B-EFG  +  Pyr.  E-ABC -{-  Pyr.  B-GCE. 

Vol.  Pyr.  B-EFG  =  \1i'b.      (1)  (XXIV.  4.) 

Vol.  Pyr.  E-ABC^  \  h  -  b'.     (2) 

Pyramids  B-GCE  and  B-ACE  have  the  same  vertex  B,  and 
their  bases  lie  in  the  same  plane  ACGE. 

.'.  Pyr.  B-GCE :  Pyr.  B-ACE : :  A  GCE :  A  ACE. 

(XXIV.  4  b.  Sch.) 

But  A  GCE :  A  ACE  ..EG:  AC.  (XIII.  1  c.  Sch. 2.) 

And  EG'.AC'.'.Vb:  Vb'.  (XVI.  2.) 

.-.  Pyr.  B-GCE :  Pyr.  B-ACE  : :  V6  :  V^. 

But  Pyr.  B-ACE  =  Pyr.  E-ABC ^^h-b'. 

.'.  Pyr.  B-GCE  :\h -b' ::^b'.  V6^. 

.-.  Vyv.  B-GCE '.\li'^V:.-s/b:l. 

[Dividing  the  consequents  by  the  common  factor  Vft^.] 


XXIV.     THE   PYI^MID   AND   THE   CONE  299 

.-.  Pyr.  B-GCE  =\h'  VW,  (3) 

.-.  volume  oiABC-G==ihb  +  ^hb'+^hVW  (adding  (1),  (2),  (3)) 

=  Lh(b  +  b'  +  Vbb'\ 

Q.E.D. 

XXIV.  6.  The  volume  of  any  frustum  equals  one 
tfdrd  the  product  of  the  altitude  into  the  sum  of  the 
two  bases  and  a  mean  proportional  hetiveen  them. 


(Let  the  student  supply  the  proof.) 


Ex.  16.  If  a,  6,  c,  etc.,  be  the  sides  of  any  section  of  a  pyramid,  and 
a',  6',  c',  etc.,  the  sides  of  any  other  section  of  the  pyramid,  then  the 
points  of  intersection  (a,  a'),  (&,  &'),  etc.,  lie  in  one  straight  line. 

Ex.  17.  The  slant  height  of  a  regular  frustum  is  20  in.  ;  its  bases  are 
squares  ;  each  side  of  the  upper  base  is  12  in.  and  each  side  of  the  lower 
base  is  8  in.     Find  the  lateral  surface  and  the  whole  surface  of  the  frustum. 

Ex.  18.  The  altitude  of  a  regular  frustum  is  8  ft.  ;  its  lower  base  is  a 
square,  each  side  of  which  is  6  ft.,  and  its  upper  base  has  an  area  of  16 
sq.  ft,    "What  is  the  volume  of  the  frustum  ? 

Ex.  19.  The  area  of  the  upper  base  of  a  frustum  is  125  sq.  yd. ;  the  area 
of  the  lower  base  is  500  sq.  yd. ;  the  altitude  is  60  yd.     Find  the  volume. 

Ex.  20.  The  altitude  of  a  regular  hexagonal  frustum  is  10  ft. ;  the 
radius  of  the  upper  base  is  6  ft.,  and  the  radius  of  the  lower  base  10  ft. 
Find  the  volume  of  the  frustum. 

Ex.  21.  Find  the  volume  of  a  regular  hexagonal  frustum  the  upper 
base  of  which  has  a  radius  of  6  ft. ;  the  lower  base  a  radius  of  10  ft. ;  and 
of  which  each  lateral  edge  is  5  ft. 


800         THE  ELEMENTS  Of  GEOMETRY 

XXIV.  7.  The  lateral  surface  of  a  regular  pyramid 
equals  one  half  the  rectangle  of  the  slant  height  and  the 
perimeter  of  the  base. 


Hyp.  If  the  slant  height  SH  of  a.  regular  pyramid  equals 
H'j  and  the  perimeter  of  the  base  equals  P, 

Cone. :  then  the  lateral  surface  =  |  -P  •  -ff'. 

Dem.     SAB,  SBC,  etc.,  are  congruent  isosceles  triangles. 

(Def.  regular  pyramid.     Cor.  (a).) 
.♦.  their  altitudes  are  each  equal  to  SH. 

Area  SCE  =  iCE-  SH.  (XIII.  2  a.) 

ATea,SCB  =  ^BC'SH. 

Similarly,  for  the  other  faces. 
.*.  by  addition, 

lateral  surface  =  ^(BC+CE-^ etc.) SH=\P'H\ 

Q.E.D. 

XXIV.  7  a.  The  lateral  surface  of  a  regular  pyramid 
equals  the  rectangle  of  the  slant  height  and  the  perim- 
eter of  a  section  parallel  to  the  base  and  m^idway  between 
the  vertex  and  the  base  {called  the  mid-section). 


XXIV.     THE   PYRAMID  AND   THE   CONE  301 

XXIV.  7  h.  The  lateral  sur- 
face of  a  regular  frustum  equals 
the  rectangle  of  the  slant  height 
and  the  perimeter  of  the  mid- 
section. 

XXIV.  7  c.  The  lateral  surface  of  a  regidar  frustum 
equals  the  rectangle  of  the  slant  height  and  half  the 
sum  of  the  perimeters  of  the  bases. 

Ex.  22.  The  lower  base  of  a  regular  pentagonal  frustum  has  a  radius 
of  a.  The  area  of  the  upper  base  is  one  third  the  area  of  the  lower.  The 
altitude  of  the  pyramid  from  which  the  frustum  is  cut  is  b.  Find  the 
convex  surface  of  the  frustum. 

Ex.  23.  The  lower  base  of  a  regular  octagonal  frustum  is  9  times  as 
large  as  the  upper.  The  radius  of  the  upper  base  is  a  and  the  slant  height 
of  the  pyramid  from  which  the  frustum  is  cut  is  b.  Pind  the  volume  of 
the  frustum. 

Ex.  24.  The  radius  of  the  upper  base  of  a  regular  octagonal  frustum  is 
12  ft.,  the  altitude  of  the  frustum  is  24  ft.  The  area  of  the  lower  base  is 
to  the  area  of  the  upper  as  16  : 9.  Find  the  convex  surface,  the  total 
surface,  and  the  volume  of  the  frustum. 

Ex.  25.  The  volume  of  a  cube  is  equal  to  the  volume  of  a  regular  tri- 
angular frustum  whose  altitude  is  c  and  the  sides  of  whose  bases  are  a  and 
6  respectively.     Find  the  length  of  the  edge  of  the  cube. 

Ex.  26.  The  total  surface  of  a  right  circular  cone,  the  radius  of  whose 
base  is  12  in.,  is  1178f  sq.  in.     Find  the  slant  height  of  the  cone. 

Ex.  27.  The  convex  surface  of  a  right  circular  cone  is  two  thirds  of  the 
total  surface.  Show  that  the  slant  height  of  the  cone  equals  the  diameter 
of  the  base. 

Ex.  28.  The  total  surface  of  a  right  circular  cone  8  ft.  in  diameter  is 
equal  to  the  total  surface  of  a  right  circular  cylinder  5  ft.  in  diameter  ;  the 
altitude  of  the  cone  is  10  ft.    Find  the  altitude  of  the  cylinder. 

Ex.  29.  Show  that  the  total  surface  of  a  right  circular  cone  equals  the 
convex  surface  of  a  cone  of  the  same  base  whose  slant  height  is  the  sum 
of  the  radius  of  the  base  and  the  slant  height  of  the  given  cone. 


802 


THE  ELEMENTS  OF  GEOMETRY 


XXIV.  8.  The  volume  of  a  prismoid  equals  one  sixth 
the  product  of  its  altitude  by  the  sum  of  the  areas  of 
the  bases  and  four  times  the  area  of  the  mid-section. 

G 


L  L 

Hyp.  If  bi  denotes  the  area  of  ABC,  b2  the  area  of  ILJ'"j 
m  the  area  of  EGH"-,  h  the  altitude,  and  V  the  volume,  of  the 
prismoid  i-C, 


Cone. :  then 


V  =  5(b,  +  b2  +  4m). 


Dem.     Join  any  point  K,  of  EGH  ">  to  the  vertices  of  the 
prismoid.     Draw  diagonals  BI,  BR,  etc.  in  each  trapezoid  face. 

Pyramid  K-ABG  has  h^  for  base  and  ~  for  altitude. 

z 

(Def.  of  mid-section.) 
61.  (XXIV.  4.) 


vol.  K-ABG  = 


Similarly, 


vol.  K-ILJ 


The  bases  of  the  other  set  of  pyramids  are  A,  cut  in  mid- 


joins  by  the  plane  MQ  of  the  mid-section. 
Let  K-IBL  be  any  one  of  these  pyramids. 

Pyramid  K-IBL  ^  Area  IBL 
Pyramid  K-FBG  ~  Area  FBG 


(XXIV.  1.) 
Draw  KG,  KF. 


^BL' 
BG' 
=  4.     (■.'BL  =  2BG.) 


(XXIV.  4  6.  Sell.) 
(XVI.  2.) 


XXIV.     THE   PYRAMID  AND   THE   CONE  303 

But  K-FBG  =  B-KFG  =  -  •  area  KFO.     (Why  ?) 

.-.  K-IBL  =  4  (K-FBG)  =  ^  •  area  KFG. 

The  sum  of  the  areas  of  the  A,  KFG,  etc.  =  m.  (Ax.  4.) 

.*.  the  sum  of  the  pyramids  whose  bases  are  the  lateral  faces 

of  the  prismoid  =  — -  m. 

.-.  F=|(6i  +  &2  +  4m). 
'  Q.E.D. 

ScH.     The  formula  just  established  is  often  called  the  Pris- 
moidal  Formula. 


Ex.  30.  The  elements  of  a  right  circular  cone  make  an  angle  of  45° 
with  the  plane  of  the  base.  Find  the  ratio  of  the  area  of  the  base  to  the 
convex  surface  of  the  cone. 

Ex.  31.  The  elements  of  a  right  circular  cone  make  an  angle  of  60° 
with  the  plane  of  the  base.  Find  the  ratio  of  the  area  of  the  base  to  the 
total  surface  of  the  cone. 

Ex.  32.  Find  the  volume  of  a  right  circular  cone  of  which  the  altitude 
is  6  ft.  and  the  base  has  a  radius  of  30  in. 

Ex.  33.  Find  the  volume  of  an  oblique  cone  of  which  the  altitude  is 
12  yd.  and  the  base  has  a  diameter  of  3  yd. 

Ex.  34.  Find  the  volume  of  a  cone  whose  base  has  a  radius  of  10  ft. 
and  whose  slant  height  is  27  ft. 

Ex.  35.  The  convex  surface  of  a  right  circular  cone  is  523.6  sq.  ft. ;  the 
slant  height  is  20  ft.    Find  the  total  surface  and  the  volume  of  the  cone. 

Ex.  36.  The  slant  height  of  a  right  circular  cone  is  60  ft. ;  each  element 
makes  an  angle  of  60°  with  the  plane  of  the  base.  Find  the  total  surface 
and  the  volume  of  the  cone. 

Ex.  37.  The  volume  of  a  right  circular  cone  the  radius  of  whose  base 
is  12  in.  is  a  cubic  foot.  Find  the  altitude  and  the  convex  surface  of  the 
cone. 

Ex.  38.  One  angle  of  a  right  triangle  is  30°.  Find  the  ratio  between 
the  volumes  of  the  cones  generated  by  revolving  the  triangle  first  about  the 
shorter  leg  as  an  axis  and  then  about  the  longer. 


804  THE   ELEMENTS  OF  GEOMETRY 

XXIV.     SUMMARY   OF   PROPOSITIONS   IN    THE    GROUP 
ON    (a)  THE  PYRAMID 

1.  If  a  set  of  lines  he  cut  hy  three  parallel  planes,  the 
lines  are  cut  proportionately. 

2.  Any  section  of  a  pyramid  parallel  to  the  hose  is 
similar  to  the  base. 

a.  The  p)erimeters  of  parallel  sections  of  a  pyramid 
are  to  each  other  {vary)  as  the  distances  of 
the  sections  from  the  vertex. 

6.  The  areas  of  two  parallel  sections  of  a  pyramid 
vary  as  the  squares  of  the  distances  of  the 
sections  from  the  vertex. 

c.  If  two  pyramids  have  equal  bases  and  equal 
altitudes,  any  sections  of  the  tivo  pyramids 
equidistant  from  the  vertices  are  equal. 

3.  Tioo  triangular  pyramids  that  have  equal  bases 
and  equal  altitudes  are  equal 

4.  The  volume  of  a  triangular  pyramid  is  one  third 
the  product  of  its  base  and  altitude. 

a.  The  volume  of  any  pyramid  equals  one  third 

the  product  of  its  base  and  altitude. 

b.  The  volumes  of  any  two  pyramids  are  to  each 

other  as   the  products   of  their   bases  and 
altitudes. 

ScH.  1.  If  the  bases  are  equal,  the  pyramids  are  to  each 
other  as  their  altitudes. 

If  the  altitudes  are  equal,  the  pyramids  are  to  each  other  as 
their  bases. 


XXIV.     THE  PYRAMID   AND  THE   CONE  305 

ScH;  2.  The  volume  of  any  polyhedron  may  be  found  by 
dividing  the  figure  into  pyramids  and  adding  the  volumes  of 
these  pyramids. 

5.  The  volume  of  a  triangular  frustum  equals  one 
third  the  product  of  the  altitude  into  the  sum  of  the 
tivo  bases  and  a  mean  proportional  between  them. 

6.  The  volume  of  any  frustum  equals  one  third  the 
product  of  the  altitude  into  the  sum  of  the  two  bases 
and  a  mean  proportional  between  them. 

7.  The  lateral  surface  of  a  regular  pyramid  equals 
one  half  the  rectangle  of  the  slant  height  and  the  perime- 
ter of  the  base. 

a.  The  lateral  surface  of  a  regular  pyramid  equals 
the  rectajigle  of  the  slant  height  and  the 
perimeter  of  a  section  parallel  to  the  base 
and  midway  betiveen  the  vertex  and  the  base 
(called  the  mid-section). 

6.  The  lateral  surface  of  a  regular  frustum  equals 
the  rectangle  of  the  slant  height  and  the 
perimeter  of  the  mid-section. 

c.  The  lateral  surface  of  a  regular  frustum  equals 
the  rectangle  of  the  slant  height  and  half  the 
sum  of  the  bases. 

8.  The  volume  of  a  prismoid  equals  one  sixth  the 
product  of  its  altitude  by  the  sum  of  the  areas  of  the 
bases  and  four  times  the  area  of  the  mid-section. 

ScH.  The  formula  just  established  is  often  called  the  Pri«- 
moidal  Formula. 


806 


THE   ELEMENTS  OF  GEOMETRY 


DEFINITIONS 
(b)   The  Conb 

A  Conical  Surface  is  a  surface  generated  by  a  straight  line 
that  moves  along  a  fixed  c\irve  and  always  passes  through  a 
fixed  point  not  coplanar  with  the  curve. 

The  moving  line  is  called  the  Generatrix. 

The  fixed  curve  is  called  the  Directrix. 

It  is  not  necessary  that  the  directrix  be  a  plane  curve.   . 

Tlie  successive  positions  of  the  generatrix  are  called  the 
Elements  of  the  Surface. 

The  fixed  point  through  which  the  generatrix  passes  is  called 
the  Vertex. 

The  Nappes  of  the  surface  are  the  two  parts  into  which  it  is 
divided  at  the  vertex. 

A  Cone  is  a  portion  of  space  bounded  by  one  nappe  of  a  coni- 
cal surface  and  a  plane  not  passing  through  the  vertex. 


The  Base  of  the  cone  is  the  plane  section  that  forms  part  of 
its  bounding  surface. 

The  Elements  of  the  Cone  are  the  segments  of  the  elements  of 
the  conical  surface  determined  by  the  vertex  and  the  base. 

The  Altitude  of  a  cone  is  the  perpendicular  from  the  vertex 
to  the  base. 

A  Circular  cone  is  one  that  has  a  circle  for  its  base. 

A  Right  cone  is  one  in  which  the  altitude  falls  at  the  center 
of  the  base. 


XXIV.     THE   PYRAMID   AND   THE   CONE  307 

The  terms  Truncated  and  Frustum  have  the  same  meaning  in 
the  ease  of  the  cone  as  in  that  of  the  pyramid. 

Note. — A  frustum  of  a  right  circular  cone  is  called  a  right  circular 
frustum. 

COEOLLARIES    FKOM    THE  DEFINITIONS 

(a)  Every  section  of  a  cone  that  passes  through  the 
vertex  is  a  triangle. 

(b)  The  right  circular  cone  is  a  cone  of  revolution,  of 
which  the  axis  is  the  altitude. 

General  Sch.  As  the  Circle  is  the  limit  of  the  Regular 
Polygon  when  the  number  of  sides  of  the  latter  is  increased 
beyond  any  assignable  value,  so  the  Circular  Cone  is  the  limit 
of  the  Pyramid  with  a  Regular  Base  when  the  number  of  lateral 
faces  is  increased  beyond  any  assignable  value. 

Hence,  whatever  propositions  ai^e  true  of  such  a  Pyramid, 
irrespective  of  the  number  of  its  lateral  faces,  are  true  of  the 
Circular  Cone. 

Accordingly,  from  the  corresponding  propositions  on  the 
pyramid,  we  have  the  following  summary : 

XXIV.     SUMMARY   OF   PROPOSITIONS   IN    THE    GROUP 
ON    (6)    THE   CIRCULAR   CONE 

9.  Sections  of  a  cone  parallel  to  the  base  are  similar 
to  the  base. 

10.  If  sections  of  a  cone  are  parallel  to  the  base, 

(a)  their  circumferences  are  to  each  other  as  their 

distances  from  the  vertex,  and 
(h)  their  areas  are  to  each  other  as  the  squares  of 

their  distances  from  the  vertex. 

11.  The  convex  surface  of  a  right  cii^cular  cone 
equals  one  half  the  product  of  the  circumference  of 
its  base  by  its  slant  height. 


808         THE  ELEMENTS  OF  GEOMETRY 

CL  If  R  is  the  radius  of  the  base  of  a  right 
circular  cone,  and  H'  the  slant  height  of  the 
cone,  the  convex  surface  of  the  cone  equals 
ttRH',  and  the  total  surface  equals  7rR(H'  +  R). 

b.  The  convex  surface  of  a  right  circidar  frustum 

equals  the  product  of  the  slant  height  by  the 
circumference  of  the  mid-section. 

c.  If  Ri  is  the  radius  of  the  upper  base  of  a  right 

circular  frustum,  Rg  the  radius  of  the  lower 
base,  and  h'  the  slant  height,  the  convex  sur- 
face of  the  frustum  equals  7rli'(Ri4-R2)- 

12.  The  volume  of  a  circular  cone  equals  one  third 
the  product  of  the  area  of  its  base  by  its  altitude. 

a.  If  R  is  the  radius  of  the  base  of  a  circular 

cone  and  H  the  altitude,  the  volume  of  the 

cone  equals  ^  ttR^H. 
h.  Any  two  cones  are  to  each  other  as  the  products 

of  the  areas  of  the  bases  by  the  altitudes. 
If  the  altitudes  are  equal,  the  cones  are  to  each 

other  as  the  areas  of  the  bases. 
If  the  bases  are  equal,  the  cones  are  to  each 

other  as  the  altitudes. 
If  the  bases  are  equal  and  also  the  altitudes,  the 

cones  are  equal, 

13.  Tlie  volume  of  the  frustum  of  a  cone  equals 
one  third  the  product  of  the  altitude  by  the  sum  of  the 
areas  of  the  bases  and  a  mean  proportional  between 
them. 


XXIV.     THE   PYRAMID   AND  THE   CONE  309 

Ex.  39.  The  radius  of  the  upper  base  of  a  right  circular  frustum  is 
25  in.  ;  the  radius  of  the  lower  base  is  36  in.  ;  the  slant  height  is  15  in. 
Find  the  convex  surface  and  also  the  total  surface  of  the  frustum. 

Ex.  40.  Show  that  the  convex  surface  of  a  right  circular  frustum 
equals  the  convex  surface  of  a  cylinder  whose  altitude  is  half  the  sum  of 
the  diameters  of  the  bases  of  the  frustum  and  whose  diameter  is  the  slant 
height  of  the  frustum. 

Ex.  41.  The  radius  of  the  upper  base  of  a  right  circular  frustum 
equals  12  ft.  ;  the  slant  height  is  14  ft.  ;  the  convex  surface  is  4928  sq.  ft. 
What  is  the  radius  of  the  lower  base  ?     (Take  ir  =  ^^^.) 

Ex.  42.  The  radii  of  the  bases  of  a  right  circular  frustum  are  16  in.  and 
24  in.  respectively.  Its  convex  surface  is  one  half  its  total  surface.  What 
is  the  slant  height  of  the  frustum  ?     What  is  the  altitude  of  the  frustum  ? 

Ex.  43.  The  convex  surface  of  a  right  circular  frustum  is  three  fourths 
of  the  convex  surface  of  the  cone  from  which  it  has  been  cut.  Find  the 
ratio  of  the  altitude  of  the  frustum  to  the  altitude  of  the  cone. 

Ex.  44.  The  radius  of  the  upper  base  of  a  frustum  is  5  ft. ;  the  radius 
of  the  lower  base  is  8  ft. ;  the  altitude  is  12  ft.  What  is  the  volume  of 
the  frustum  ? 

Ex.  45.  The  radii  of  the  bases  of  a  right  circular  frustum  are  20  ft.  and 
24  ft.  respectively.    Its  volume  is  4928  cu.  ft.     What  is  its  altitude  ? 

Ex.  46.  The  volume  of  a  right  circular  frustum  is  2200  cu.  ft. ;  the 
slant  height  is  13  ft.  and  the  altitude  is  12  ft.     Find  the  radii  of  the  bases. 

Ex.  47.  How  much  sheet  tin  will  be  required  to  construct  a  water  pail 
18  in.  in  diameter  at  the  top,  14  in.  in  diameter  at  the  bottom,  and  having 
a  capacity  of  8  gal.,  allowing  a  waste  of  10  per  cent -in  the  material  in 
seams  and  cuttings  ? 

Ex.  48.  The  volume  of  a  right  circular  frustum  is  three  fourths  of  the 
volume  of  the  cone  from  which  it  has  been  cut.  What  is  the  ratio  of  the 
altitude  of  the  frustum  to  the  altitude  of  the  cone  ? 

Ex.  49.  The  area  of  the  lower  base  of  a  right  circular  frustum  is  four 
times  the  area  of  the  upper  base.  The  volume  of  the  frustum  equals  the 
volume  of  a  right  circular  cylinder  whose  base  is  the  upper  base  of  the 
frustum.  Find  the  ratio  of  the  altitude  of  the  cylinder  to  the  altitude  of 
the  frustum. 

Ex.  50.   From  a  right  circular  frustum  whose  upper  

diameter  is  2  a  and  lower  diameter  4  a  is  bored  out, 
as  shown  in  the  figure,  a  right  circular  cone  whose 
base  has  a  diameter  2  a.  Find  the  ratio  of  the  volume 
of  the  original  frustum  to  the  volume  of  the  hollow 
frustum. 


810 


THE  ELEMENTS  OF  GEOMETRY 


A  •"  G  is  a  prismoid ;  MIDS  is  a  mid-section  parallel  to  the  bases. 
/l,  any  point  whatever  of  MIBS\  is  joined  to  the  vertices  of  the  prismoid. 

Denote  the  altitude  by  h,  the  upper  base  by  6,  the  lower  by  6',  and  the 
mid-section  by  m. 

Ex.  51.   Show  that  the 
volume  of  pyramid 

K-ABCT=h, 
6 

and  the  volume  of  pyramid 

K-EFGIi=-b'. 

Draw  KH  ±  DI  and 
OJJLhoWi  CB  And  KH. 
.Draw  OQ  through  H 

and  KL±OQ.  , 

(XXI.  Prob.  2.)    ^         •  ^ 

OQ  ±  DI  and  KL  ±  BCGF.  (XXI.  5,  5  a,  5  6,  3,  3  a.) 

Ex.  52.  Show  that  OJ-  KH=^  KL-  OH, 

whence,  2  OJ  •  KH  -  DI=KL'20H.  DI, 

and  the  volume  of  pyramid  K-CBFG  -\KL'  OQ-DI 


2l\DKI 
^ADKI. 


Ex.  53.   Hence  show,  by  considering  all  the  pyramids,  that  the 

volume  of  the  prismoid  =  -  (6  +  &'  +  4  w). 
6 

Ex.  54.  Show  that  if  the  prismoid  be  the  frustum  of  a  pyramid,  the 
above  expression  may  be  reduced  to  that  given  in  XXIV.  6. 

Ex.  55.  Show  that  any  plane  through  the  center  of  a  parallelepiped 
divides  the  figure  into  two  equal  parts. 

Ex.  56.    Divide  a  parallelepiped  into  two  equal  parts  by  a  right  section. 

Ex.  57.  Divide  a  parallelepiped  into  two  equal  parts  by  a  plane  parallel 
to  a  given  plane. 

Ex.  58.  A  parallelepiped  is  given  and  also  two  lines  in  space.  Show 
how  to  pass  a  plane  parallel  to  the  given  lines  that  shall  divide  the  paral- 
lelepiped into  two  equal  parts. 


m 


XXV.     GROUP   ON   THE   SPHERE 

DEFINITIONS 

A  Spherical  Surface  is  a  surface  generated  by  the  revolution 
of  a  semicircumference  (the  generatrix)  about  its  diameter  as 
an  axis. 

A  Sphere  is  a  portion  of  space  inclosed  by  a  spherical  surface. 

Note.  — As  in  the  case  of  circle  and  circumference,  the  terms  sphere 
and  spherical  surface  are  used  interchangeably  where  no  confusion  is 
likely  to  result. 

The  Radius  of  the  generatrix  is  the  radius  of  the  sphere. 

A  Great  Circle  of  a  sphere  is  a  circle  on  the  sphere  whose 
plane  passes  through  the  center  of  the  sphere. 

A  Small  Circle  of  a  sphere  is  a  circle  on  the  sphere  whose 
plane  does  not  pass  through  the  center  of  the  sphere. 

The  Axis  of  a  Circle  of  a  sphere  is  the  perpendicular  to  the 
plane  of  the  circle  at  its  center. 

The  Poles  of  a  circle  are  the  points  in  which  the  axis  of  the 
circle  intersects  the  surface  of  the  sphere. 

The  Polar  Distance  of  a  point  on  a  circumference  on  the 
sphere  is  the  length  of  the  great  circle  arc  joining  the  point  to 
the  nearer  pole  of  that  circumference. 

A  Plane  is  Tangent  to  a  sphere  when  it  has  one  point,  and 
only  one,  in  common  with  the  surface  of  the  sphere. 

A  sphere  is  Circumscribed  to  a  polyhedron  when  all  the  ver- 
tices of  the  polyhedron  lie  on  the  surface  of  the  sphere. 

A  sphere  is  Inscribed  in  a  polyhedron  when  the  faces  of  the 
polyhedron  are  all  tangent  to  the  sphere. 

A  Zone  is  a  portion  of  the  surface  of  the  sphere  comprised 
between  two  parallel  planes. 

311 


812  THE  ELEMENTS  OF  GEOMETRY 

A  Spherical  Segment  is  a  portion  of  the  volume  of  the  sphere 
comprised  between  two  parallel  planes. 

A  Spherical  Sector  is  the  portion  of  the  sphere  generated  by 
the  revolution  of  a  sector  of  the  generatrix. 


PROPOSITIONS 

XXV.  1.  Every  plane  section  of  a  sphere  is  a  circle. 


.  F 

Hjrp.    If  the  plane  MQ  cuts  the  sphere  S  in  the  line  ABC  •••, 

Cone. :  then  ABC  •••  is  a  circle. 

Dem.    Through  S  draw  PP  perpendicular  to  plane  MQ,  and 
intersecting  the  plane  MQ  in  K. 

Suppose  T  to  be  any  point  whatever  in  ABC  •••. 

Draw /ST  and /i'r. 

Let  ST=Ry  KT=r,  and  SK=d. 

Then,  in  the  right  triangle  SKT, 


r^VR'-d\  (XIV.  la.) 

But  R  is  constant  for  the  given  sphere,  andd  is  constant  for 
the  given  plane. 

.*.  r  is  constant  for  all  points  in  their  line  of  intersection. 
"That  is,  the  distance  of  T  from  K' is  the  same  for  all  posi- 
tions of  T. 

-         .-.  ABC  •••  is  a  6if6le"  whose  center  is  h": 

-  .        .     .  ;  OLD 


♦'..,-v4l 


XXV.     THE  SPHERE  813 

XXV.  1  a.  The  join  of  the  center  of  the  sphere  and 
the  center  of  any  circle  on  the  sphere  is  the  axis  of  the 
circle,  and  conversely, 

XXV.  1  K  The  locus  of  the  centers  of  all  spheres  that 
pass  through  three  given  points  is  the  axis  of  the  circle 
that  passes  through  the  points, 

XXV.  1  c.  Circles  cut  out  by  planes  equidistant  from 
the  center  of  the  sphere  are  equal,  and  conversely. 


Dem.  T  =  Vi2'  -  d\  (XIV.  1  a.) 

.-.  if  d  remains  the  same,  r  must  remain  the  same,  and  if  r 

remains  the  same,  d  must  remain  the  same. 

Q.E.D. 

XXV.  Id.  Of  two  circles  cut  by  planes  unequally 
distant  from  the  center,  the  one  nearer  the  center  is  the 
greater,  and  conversely. 


Dem.  r  =  Vi2'  -  d\  (XIV.  1  a.) 


When  d  =  R,r=^B'-R^  =  0. 

As  d  diminishes,  R^  —  d^  increases. 


.-.  r,  which  =  -vR^  —  d^,  increases. 

.*.  the  nearer  a  circle  is  to  the  sphere  center,  the  greater  is 

the  radius  of  the  circle ;  i.e.  the  greater  is  the  circle. 

Q.E.D. 

XXV.  1  e.  The  polar  distances  of  all  points  in  the 
circumference  of  a  circle  of  the  sphere  are  equal. 

ScH.  From  this  property  the  polar  distance  of  the  points  of 
a  circle  are  called  arc-radii  of  the  circle. 

The  arc-radius  of  a  great  circle  is  a  great-circle  quadrant, 
called  simply  a  quadrant. 

The  arc-radius  of  a  small  circle  is  less  than  a  quadrant. 


814 


THE  ELEMENTS  OF  GEOMETRY 


XXV.  1/  Three  points  on  the  sphere  surface  {not  on 
the  same  great  circle,  no  two  of  which  are  the  extremities 
of  a  diameter)  are  necessary  and  sufficient  to  determine 
a  small  circle  of  the  sphere. 

XXV.  1  g.  Two  points  on  the  sphere  surface  (not  the 
extremities  of  a  diameter)  are  necessary  and  sufficient 
to  determine  a  great  circle  of  the  sphere, 

XXV.  2.  A  plane  perpendicular  to  a  radius  at  its 
extremity  is  tangent  to  the  sphere,  and  conversely. 


Hyp.     If  the  plane  MQ  is  perpendicular  to  /Sr  at  T, 

Cone:  then  plane  MQ  is  tangent  to  the  sphere;  and  con- 
versely. 

Dem.  Let  DL  be  any  plane  perpendicular  to  ST^  and 
cutting  the  sphere. 

All  points  common  to  DL  and  the  sphere  lie  in  the  circle 
ABC,  of  radius  r  =  -^B?-^.  (XIV.  1  a.) 

Move  the  plane  DL  away  from  S,  keeping  it  ±  to  ST. 

As  d  increases,  r  decreases ;  until,  when 

%,e.  the  circle  ABC  which  contains  all  points  common  to  the 
plane  and  the  sphere  becomes  itself  a  point,  T. 

.'.  this  point  is  a  unique  point  common  to  the  sphere  and 
the  plane  DL,  i.e.  DL  is  tangent  to  the  sphere. 

(Def.  of  tangent  plane.) 


XXV.     THE   SPHERE 


315 


But  when  d  =  R,  the  planes  MQ  and  DL  are  both  perpen- 
dicular to /ST  at  T. 

.'.  plane  MQ  is  identical  with  plane  DL.  (XXI.  3  a.) 

.-.  plane  MQ  is  tangent  to  the  sphere  at  T,  and  is  unique. 

Q.E.D. 

XXV.  3.    Through  any  four  poi7its,  not  in  the  same 
plane,  a  unique  sphere  may  he  passed. 


TLyp.     If  A,  B,  C,  E,  are  not  in  the  same  plane, 

Cone. :  then  a  unique  sphere  may  be  passed  through  A,  B, 
C,  and  E. 

Dem.     No  three  of  the  points  can  be  collinear. 

(Def.  of  plane,  g,  XXI.) 

The  locus  of  the  centers  of  spheres  through  A,  B,  and  G 
is  KL,  perpendicular  to  plane  ABC  at  K,  the  center  of  the 
circle  through  A,  B,  and  O.  (XXV.  1  b.) 

All  points  equidistant  from  A  and  E  lie  on  the  plane  MQ 
perpendicular  to  AE.  (XXI.  3  b.) 

This  plane  must  cut  KL,  as  at  S.  (Why  ?) 

.-.  S  is  equidistant  from  A,  B,  C,  and  E.  (Why  ?) 

.*.  a  sphere  with  S  as  center  and  SA  as  radius  passes 
through  A,  B,  C,  and  E. 

Again,  the  point  S  is  unique.  (Why  ?) 

.*.  the  sphere  through  A,  B,  C,  and  E  is  unique. 

Q.E.D. 


S16 


THE  ELEMENTS  OF  GEOMETRY 


XXV.  3  a.  The  perpendiculars  to  the  faces  of  a  tetra- 
hedron at  their  circumcenters  are  concurrent. 

Outline  Dem.  Draw  AB,  BC,  etc.,  forming  the  tetrahedron 
ABCE. 

Then  K  is  the  circum center  of  triangle  ABC,  and  KL  the 
axis  of  the  circle  through  A,  B,  and  C. 

Similarly  with  each  of  the  other  faces  of  ABCE. 

.'.  the  perpendiculars  concur  at  the  center  of  the  circum- 
sphere.  (XXV.  1  a.) 

Q.E.D. 

XXV.  3  b.  The  six  planes  mid-normal  to  the  edges  of 
a  tetrahedron  have  a  unique  point  in  common. 

XXV.  ^.  If  R  is  the  radius  of  a  sphere,  the  area  of 
the  surface  of  the  sphere  equals  4  ttRK 


■ 

p 

A'    ^ 

^R^ 

V 

^w 

M 

B' 

1 

\ 

> 

C      i 

D 

^      -4 

L 

,J 

Hyp.  If  R  is  the  radius  of  the  sphere  generated  by  the 
semicircle  HBL, 

Cone. :  then  the  area  of  the  surface  is  4  vR^. 

Dem.  Let  AB  be  a  side  of  a  regular  n-gon  inscribed  in  the 
circle,  A'B'  its  projection  on  the  axis  HL,  and  M'  the  projection 
of  its  middle  point  M. 


XXV.     THE   SPHERE  317 

Draw  AG  parallel  to  HL,  and  draw  MS'. 

The  surface  generated   by  AB  when  HBL  generates  the 

sphere,  or 

Surf.  AB=2  ttMM'  •  AB. 

[The  convex  surface  of  the  frustum,  etc.  (XXIV.  11  h.).] 

But  rt.  A  MM'S'  ^  rt.  A  ABG.  (XV.  Exs.  50,  51.) 

.-.  AB :  S'M=:  AG{=  A'B')  :  MM'. 

.-.  AB'MM'=S'M'A'B'. 

.-.  Surf.  AB  =  27rS'M  •  A'B'. 

Similarly  for  the  surface  generated  by  HA,  BC,  etc. 
.-.  the  surface  generated  by  the  semi-polygon 

HAB ...  Z  =  2  7rS'M(HA'  +  A'B'  +  B'C  +  •••)  =  2  ttS'M-  HL 
=  2 7rS'M'2 R  =4:7rR-  S'M. 

But  if  n  be  indefinitely  increased,  the  semi-polygon  =  the 
generatrix,  and  the  surface  generated  by  the  semi-polygon 
=  the  sphere  surface  and 

S'M^R. 

.*.  the  sphere  surface  =  4  ttR  .  i?  =  4  irR^. 

Q.E.D. 

XXV.  ^a.  If  h  is  the  altitude  of  a  zone  on  a  sphere 
of  radius  E,  the  area  of  the  zone  equals  2  irRli. 

Outline  Dem.  The  surface  generated  by  any  portion  of  the 
semi-polygon,  e.g.  ABC,  is  2  ttR  •  A'C.  At  the  limit,  this  sur- 
face becomes  a  zone  of  altitude  A'C  =  li. 

Ex.  1.  Find  the  surface  and  the  vohime  of  a  sphere  whose  radius  is 
10  ft. 

Ex.  2.   Find  the  total  surface  of  a  hemisphere  whose  radius  is  12  ft. 

Ex.  3.    Tlie  surface  of  a  sphere  is  1256.64  sq.  ft.     What  is  the  radius  ? 

Ex.  4.  Show  that  the  area  of  the  surface  of  a  sphere  is  the  same  as  that 
of  a  circle  wliose  radius  is  the  diameter  of  the  sphere. 

Ex.  5.  Find  the  area  of  a  zone  whose  altitude  is  10  ft.  and  which  is 
situated  on  a  sphere  of  20  ft.  radius. 


818 


THE   ELEMENTS   OF   GEOMETRY 


XXV.  5.  If  E  is  the  radius  of  a  sphere^  the  volume 
of  the  sphere  is  |  nli^. 


Hyp.    If  B 

is  the  length 
of  the  radius 
of  the  sphere 

s, 


Cone. :  then  the  volume  of  the  sphere  S  is  ^  irB?. 
Dam.     Circumscribe  the  cube  P^  about  S. 
Draw  SA,  SB,  etc.,  thus  dividing  P^  into  pyramids. 
Volume  of  pyramid  S-ABCE  =  ABCE  -  \  R.      (XXIV.  4  a.) 
Similarly,  for  the  other  pyramids  of  Pj. 

.-.  volume  of  P^  —  (surface  of  Pj)  •  \  R. 

At  Q,  •••,  where  SA,  etc.,  cut  the  surface  of  S,  pass  planes 
tangent  to  S  and  truncating  (cutting  off  the  corners)  symmet- 
rically the  cube  Pi. 

Denote  the  polyhedron  thus  obtained  by  Pg. 

Draw  SL,  SO,  etc.,  dividing  Pa  into  pyramids. 

Volume  of  pyramid  S-LMO  =  LMO  -  i  R,     (XXIV.  4.) 
Similarly,  for  the  other  pyramids  of  Pg. 

.%  volume  of  P^,  =  (surface  of  P^  •  J  -B. 
By  passing  a  new  set  of  tangent  planes  truncating  Pg,  we 
obtain  a  new  circumscribed  polyhedron,  Pg,  such  that 
volume  of  Pg  =  (surface  of  Pg)  •  \R. 
This  process,  may  be  continued  indefinitely. 
But  each  successive  polyhedron  is  nearer  in  volume  to  the 
sphere  than  the  preceding. 

Again,  the  volume  of  any  polyhedron  thus  obtained  will  be 
greater  than  the  volume  of  S,  since  its  vertices  are  without  S, 
and  its  faces  are  tangent  planes. 


XXV.     THE   SPHERE  319 

But  the  excess  of  volume  may  be  made  as  small  as  we  please 
by  continuing  the  truncation  far  enough. 
.%  if  P  be  any  of  the  polyhedra, 

volume  of  P  =  volume  of  S. 
But  volume  of  P=  (surface  of  P)  •  |  R. 

.',  volume  oi  S  =  (surface  oi  S)  •  \E 

=  4:7rR''iE  (XXV.  4.) 

Q.E.D. 

XXV.  5  a.  The  volume  of  a  spherical  sector  equals 
one  third  the  product  of  the  zone  that  forms  its  base 
and  the  radius  of  the  sphere. 

Ex.  6.  The  surface  of  a  sphere  is  equal  to  the  surface  of  a  cube.  Find 
the  ratio  of  the  diameter  of  the  sphere  to  the  edge  of  the  cube. 

Assuming  the  radius  of  the  earth  to  be  3967.2  miles,  and  the  chord  of 
an  arc  of  23 1°  to  be  .407  of  the  radius,  find : 
Ex.  7.   The  area  of  each  Frigid  Zone. 
Ex.  8.   The  area  of  each  Temperate  Zone. 
Ex.  9.   The  area  of  the  Torrid  Zone. 

Ex.  10.  The  total  surface  of  a  cylinder  of  revolution  is  equal  to  the 
total  surface  of  a  hemisphere.  Find  the  ratio  of  the  two  volumes,  if  the 
diameter  of  the  cylinder  equals  its  altitude. 

Ex.  11.  The  total  surface  of  a  cone  of  which  the  altitude  equals  the 
radius  of  the  base  is  equal  to  the  surface  of  a  sphere.  Find  the  ratio  of 
the  altitude  of  the  cone  to  the  diameter  of  the  sphere. 

Ex.  12.  Show  that  if  spheres  be  described  on  the  sides  of  a  right 
triangle  as  diameters  the  surface  of  the  sphere  on  the  hypotenuse  will  be 
equal  to  the  sum  of  the  surfaces  of  the  other  two  spheres. 

Ex.  13.  The  radii  of  two  spheres  are  a  and  b.  Show  how  to  construct 
the  sphere  whose  area  is  equal  to  the  sum  of  the  areas  of  the  given 
spheres. 

Ex.  14.  Show  that  if  spheres  be  described  on  three  concurrent  edges  of 
a  rectangular  parallelepiped  as  diameters  the  sum  of  the  surfaces  of  the 
spheres  will  equal  the  surface  of  the  sphere  described  on  the  diagonal  of 
the  parallelepiped  as  diameter. 


820  THE  ELEMENTS  OF  GEOMETRY 

XXV.     SUMMART    OP    PROPOSITIONS    IN    THE    GROUP 
ON   THE   SPHERE 

1 .  Every  plane  section  of  a  sphere  is  a  circle. 

a.  Tliejoin  of  the  center  of  a  sphere  and  the  center 

of  any  circle  on  the  sphere  is  the  axis  of  this 
circle,  and  conversely. 

b.  The  locus  of  the  centers  of  all  spheres  that  pass 

through  three  given  points  is  the  axis  of  the 
circle  that  2^cisses  through  the  points. 

c.  Circles  cut  ojf  by  planes  equidistant  from  the 

center  of  the  sphere  are  equal,  and  conversely. 

d.  Of  tioo  circles  cut  by  planes  unequally  distant 

from  the  center  of  the  sphere,  the  one  nearer 
the  center  is  the  greater,  and  conversely. 

e.  The  polar  distances  of  all  points  in  the  circum- 

ference of  a  circle  of  the  sphere  are  equal. 

ScH.  From  this  property  the  polar  distances  of  the  points 
of  a  cirple  are  called  arc-radii  of  the  circle. 

The  arc-radius  of  a  great  circle  is  a  great-circle  quadrant, 
called  simply  a  quadrant. 

The  arc-radius  of  a  small  circle  is  less  than  a  quadrant. 

/  Three  points  on  the  sphere  surface  {not  on  the 
same  great  circle,  and  no  two  of  which  are 
the  extremities  of  a  diameter)  are  necessary 
and  siifficient  to  determiiie  a  small  circle  of 
the  sphere. 

g.  Tivo  points  on  the  sphere  surface  {not  the  extremi- 
ties of  a  diameter)  are  necessary  and  sufficient 
to  determine  a  great  circle  of  the  sphere. 


XXV.     THE   SPHERE  321 

2.  A  plane  perpendicular  to  a  radius  at  its  extremity 
is  tangent  to  the  sphere,  and  conversely. 

3.  Through  any  four  points  not  in  the  same  plane 
a  unique  sjjhere  may  he  passed. 

a.  The  p)erpendiculars  to  the  faces  of  a  tetrahedron 

erected  at  their  circumcentevs  are  concurrent. 

b.  TJie  six  planes  mid-normal  to  the  edges  of  a 

tetrahedron  have  a  unique  point  in  common. 

A.  If  R  is  the  radius  of  a  sphere,  the  area  of  the  sur- 
face of  the  sphere  is  4  ttR^. 

a.  If  h.  is  the  altitude  of  a  zone  on  a  sphere  of 
radius  R,  the  area  of  the  zone  equals  2'n-Rli. 

5.  If  R  is  the  radius  of  a  sphere,  the  volume  of  the 
sphere  equals  |  irR^. 

cu  The  volume  of  a  spherical  sector  equals  one 
third  the  product  of  the  zone  that  forms  its 
base  and  the  radius  of  the  sphere. 


Ex.  15.  Find  the  volume  of  a  sphere  that  will  just  fit  into  a  cubical  box 
each  edge  of  the  inside  measurement  of  which  is  10  in. 

Ex.  16.  Find,  to  within  .01  ft.,  the  edge  of  a  cube  whose  volume 
equals  the  volume  of  a  sphere  of  radius  12  ft. 

Ex.  17.    The  volume  of  a  sphere  is  7241}  cu.  ft.    What  is  the  radius  ? 

Ex.  18.  The  number  that  expresses  the  volume  of  a  certain  sphere  is 
the  same  as  the  number  that  expresses  the  surface  of  the  sphere.  Find 
the  radius  of  the  sphere. 

Ex.  19.  A  cube  and  a  sphere  have  the  same  surface.  Which  tas  the 
greater  volume  ?    Prove. 


822 


THE  ELEMENTS  OF  GEOMETRY 


PROBLEMS 

XXV.  Prob.  1.   To  find  the  diameter  of  a  material 
sphere. 

G' 


Given.     The  material  sphere  AB  •••  i?. 
Required.     The  diameter  of  this  sphere. 

Const.  With  any  point,  as  Aj  as  a  pole,  and  any  convenient 
opening  of  the  dividers,  describe  a  circle,  B---  C. 

In  this  circle  select  any  3  points,  F,  E,  6r,  and  with  the 
dividers  set  off  E'F'  =  EF,  F'G'  =  FG,  and  G'E'  =  GE  so  as 
to  form  the  triangle  E'F'G'  congruent  to  triangle  EFG. 

Find  the  circum-radius  of  this  triangle,  K'E',  which  will  also 
be  the  circum-radius  of  EFG,  i.e.  the  radius  of  circle  B  >"  C. 

Construct  rt.  A^i^J^f  with  hypotenuse  HM  =  chord  AE  and 
\egLM=K'E'. 

Erect  a  perpendicular  to  HM  3it  M,  and  extend  this  perpen- 
dicular to  meet  HL,  say  at  Q. 

HQ  is  the  diameter  required. 

Proof.     Suppose  the  diameter  AR  drawn ;  also  ACy  CR. 

rt.  A  HLM  ^  rt.  A  ARE.  (Const.) 

.-.  In  vtAACR,  HMQ, 

^A  =  ZH.  (Hom.Zs^A.) 

Z  AdR  =  Z.  HMQ,  (Both  rt.  A) 


XXV.     THE   SPHERE  323 

AC=HM.  '  (Const.) 

.-.  rt.  A  ACli  ^  rt.  A  HMQ.  (V.  2.) 

.-.  AB  =  HQ.  (Horn.  s's.  -^  A.) 

Q.E.D. 

Ex.  20.  An  iron  cannon  ball  12  in.  in  diameter  weighs  225  lbs.  What 
is  the  diameter  of  a  ball  of  the  same  material  weighing  1800  lbs.  ? 

Ex.  21.  The  volume  of  a  sphere  equals  the  volume  of  a  cube.  Find 
the  ratio  of  the  diameter  of  the  sphere  to  the  edge  of  the  cube. 

Find  also  the  ratio  of  the  diameter  of  the  sphere  to  the  diagonal  of  the 
cube. 

Ex.  22.  The  volume  of  a  sphere  is  equal  to  the  volume  of  a  cone  whose 
slant  height  is  double  the  radius  of  its  base.  Find  the  ratio  of  the  total 
surfaces  of  the  two  figures. 

Ex.  23.  The  sura  of  the  surfaces  of  three  spheres  is  equal  to  a  circle  of 
which  the  radius  is  twice  the  diagonal  of  an  oblong  block  whose  edges  are 
a,  6,  and  c.  The  volumes  of  the  three  spheres  are  in  the  ratio  of  a^  :b^  :  c^ 
Find  the  radii  of  the  spheres. 

Ex.  24.  A  sphere  just  fits  into  a  regular  triangular  prism  each  base 
edge  of  which  is  a.    Find  the  volume  of  the  sphere. 

Ex.  25.  The  surface  of  a  cube  equals  the  surface  of  a  sphere.  Find 
the  ratio  of  the  volume  of  the  cube  to  the  volume  of  the  sphere. 

Ex.  26.  A  sphere  and  a  regular  tetrahedron  have  the  same  surface. 
Find  the  ratio  of  their  volumes. 

Find  the  ratio : 

Ex.  27.  Of  the  surface  of  a  sphere  to  the  surface  of  the  circumscribed 
cube. 

Ex.  28.   Of  the  surface  of  a  sphere  to  the  surface  of  the  inscribed  cube. 

Ex.  29.  Of  the  volume  of  a  sphere  to  the  volume  of  the  circumscribed 
cube. 

Ex.  30.   Of  the  volume  of  a  sphere  to  the  volume  of  the  inscribed  cube. 

Def.  A  Principal  Section  of  a  Surface  of  Revolution  is  a  section  that 
passes  through  the  axis. 

The  principal  sections  of  a  certain  right  circular  cylinder  are  squares 
each  side  of  which  is  2  a. 

Find  the  ratio ;  ♦ 

Ex.  31.  Of  the  convex  surface  of  the  cylinder  to  the  surface  of  the 
inscribed  sphere. 


824         THE  ELEMENTS  OF  GEOMETRY 

Ex.  32.  Of  the  total  surface  of  the  cylinder  to  the  surface  of  the 
inscribed  sphere. 

Ex.  33.  Of  the  volume  of  the  cylinder  to  the  volume  of  the  inscribed 
sphere. 

Note.  —  The  three  preceding  problems  were  first  solved  by  Archimedes. 

Ex.  34.  Of  the  convex  surface  of  the  cylinder  to  the  surface  of  the 
circumscribed  sphere. 

Ex.  35.  Of  the  total  surface  of  the  cylinder  fb  the  surface  of  the  cir- 
cumscribed sphere. 

Ex.  36.  Of  the  volume  of  the  cylinder  to  the  volume  of  the  circum- 
scribed sphere. 

Ex.  37.  Of  the  volume  of  the  circumscribed  sphere  to  the  volume  of  the 
inscribed  sphere. 

The  vertex  angle  of  a  right  circular  cone  is  60° ;  its  slant  height  is  2  a. 

Find  the  ratio : 

Ex.  38.  Of  the  convex  surface  of  the  cone  to  the  surface  of  a  sphere 
of  radius  a. 

Ex.  39.  Of  the  total  surface  of  the  cone  to  the  surface  of  the  sphere 
of  radius  2  a. 

Ex.  40.  Of  the  total  surface  of  the  cone  to  the  surface  of  a  sphere 
whose  radius  is  the  altitude  of  the  cone. 

Ex.  41.  Of  the  volume  of  the  cone  to  the  volume  of  the  sphere  of 
radius  a. 

Ex.  42.  Of  the  volume  of  the  cone  to  the  volume  of  the  sphere  whose 
radius  is  the  altitude  of  the  cone. 

Ex.  43.  Of  the  volume  of  the  cone  to  the  volume  of  the  sphere  whose 
surface  equals  the  convex  surface  of  the  cone. 

Ex.  44.  Of  the  volume  of  the  cone  to  the  volume  of  the  sphere  whose 
surface  equals  the  total  surface  of  the  cone. 

The  surface  of  a  sphere  equals  the  total  surface  of  a  right  circular  cone 
whose  principal  sections  are  equilateral  triangles ;  and  also  equals  the  total 
surface  of  a  right  circular  cylinder  whose  principal  sections  are  squares. 

Find  the  ratio : 

Ex.  45.    Of  the  volum^of  the  sphere  to  the  volume  of  the  cylinder. 
Ex.  46.    Of  the  volume  of  the  sphere  to  the  volume  of  the  cone. 
Ex.  47.   Of  the  volume  of  the  cylinder  to  the  volume  of  the  cone. 


XXV.     THE   SPHERE  325 

Ex.  48.  Of  the  convex  surface  of  the  cylinder  to  the  convex  surface  ot 
the  cone. 

Ex.  49.  An  iron  sphere  whose  radius  is  8  in.  is  melted  and  recast  in 
the  form  of  a  hollow  right  circular  cylinder  whose  altitude  and  interior 
diameter  are  each  0  in.     Eind  the  thickness  of  the  cylinder. 

Ex.  50.  Find  the  weight  of  a  hollow  spherical  cast-iron  shell  whose 
exterior  diameter  is  20  in.  and  whose  interior  diameter  is  17  in. 

Ex.  51,  A  hollow  spherical  shell  has  a  capacity  of  2  gal.  ;  its  exterior 
diameter  is  12  in.     Find  the  thickness  of  the  shell. 

Ex.  52.  Find  the  locus  of  the  center  of  a  sphere  10  in.  in  diameter,  the 
surface  of  which  passes  throutih  a  given  point  A. 

What  is  the  locus  of  a  point  at  a  constant  distance  (d)  from  a  given 
point  ? 

What  is  the  locus  : 

Ex.  53.  Of  the  center  of  a  sphere  passing  through  two  given  points, 
A  and  B  ? 

Ex.  54.  Of  the  center  of  a  sphere  of  radius  r  whose  surface  passes 
through  two  given  points,  A  and  B  ? 

Ex.  55.  Of  the  center  of  a  sphere  whose  surface  passes  through  three 
given  points,  A,  B,  and  C  ? 

Ex.  56.  Of  the  centers  of  spheres  whose  surfaces  all  contain  a  given 
circle  ? 

What  is  the  locus  of  the  center  of  a  sphere : 

Ex.  57.  Tangent  to  three  given  planes  ? 

Ex.  58.  Tangent  to  three  intersecting  lines  ? 

Ex.  59.  Tangent  to  a  given  plane  at  a  given  point  ? 

Ex.  60.  Tangent  to  a  given  cylindrical  surface  at  a  given  point  ? 

Ex.  61.  Tangent  to  a  given  sphere  at  a  given  point  ? 

Ex.  62.  Tangent  to  a  cone  of  revolution  at  a  given  point  ? 

Ex.  63.  Tangent  to  two  concentric  spheres  ? 

Find  the  locus  of  the  center  of  a  sphere  of  given  radius  r  that  satisfies 
the  conditions  that  follow : 

Ex.  64.  Tangent  to  a  given  plane. 

Ex.  65.  Tangent  to  two  intersecting  planes. 

Ex.  66.  Tangent  to  a  cylindrical  surface  of  revolution. 

Ex.  67.  Tangent  to  a  conical  surface  of  revolution. 


826         THE  ELEMENTS  OF  GEOMETRY 

Ex.  68.   Tangent  to  a  sphere  of  radius  m. 

Ex.  69.  Tangent  to  two  cylindrical  surfaces  of  revolution,  of  diameters 
a  and  b. 

Ex.  70.   Tangent  to  two  spheres  of  equal  radius  a. 

Ex.  71.  To  find  a  point  equidistant  from  three  given  points  and  also 
equidistant  from  two  other  points  not  in  the  plane  of  the  first  three. 

To  construct  a  sphere  of  radius  r : 

Ex.  72.  That  shall  pass  through  a  given  point  and  be  tangent  to  two 
given  planes. 

Ex.  73.  That  shall  pass  through  two  given  points  and  shall  also  be 
tangent  to  a  given  sphere. 

Ex.  74.  That  shall  pass  through  a  given  point  and  shall  also  be  tangent 
to  a  cylinder  of  revolution  along  a  given  element. 

Ex.  75.  That  shall  pass  through  two  given  points  and  shall  also  be 
tangent  to  a  cylinder  of  revolution  along  a  given  element. 

Ex.  76.  That  shall  pass  through  a  given  point  and  be  tangent  to  a  given 
sphere  and  a  given  plane. 

Ex.  77.    Tangent  to  three  given  planes. 

Ex.  78.    Tangent  to  three  given  concurrent  lines. 

Ex.  79.   That  shall  pa.ss  through  three  given  points. 

Ex.  80.  Tangent  to  two  intersecting  given  lines  and  passing  through  a 
given  point. 

Ex.  81.  Tangent  to  two  non-intersecting  lines  and  passing  through  a 
given  point. 

Ex.  82.  Tangent  to  two  given  spheres  of  radii  &  and  c,  respectively. 
and  passing  through  a  given  point. 

Show  that,  given  a  point  and  a  sphere, 

Ex.  83.  In  general,  an  infinite  number  of  tangent  lines  can  be  drawn 
through  the  given  point  tangent  to  the  given  sphere. 

Ex.  84.  These  tangents  either  lie  in  the  same  plane  or  form  th# 
elements  of  a  cone. 

Ex.  85.   The  cone  of  tangents  is  a  cone  of  revolution. 

Ex.  86.  In  general,  an  infinite  number  of  planes  may  be  passed  through 
the  point  tangent  to  the  sphere. 

Ex.  87.  These  planes  will  each  be  tangent  to  the  cone  of  tangent  lines 
through  the  point. 

Ex.  88.  When  will  it  be  impossible  to  pass  more  than  one  tangent 
plane  through  the  point  ? 


XXVo     THE   SPHERE 


327 


Ex.  89.  When  will  it  be  impossible  to  pass  either  a  tangent  line  or  a 
tangent  plane  through  the  point  ? 

To  pass  a  plane  tangent  to  a  given  sphere, 

Ex.  90.    At  a  point  on  the  sphere. 

Ex.  91.   Through  a  point  without  the  sphere. 

Ex.  92.  If  AB  and  AEare  tangent  to  the  sphere 
S,  show  that  ABS  is  a  right  triangle.  Show  further 
that  AB^  =  A0-  AS.  Hence,  show  how  to  find  FC 
when  AF  and  the  radius  SB  are  given. 

Ex.  93.  An  electric  light  is  placed  at  ^,  3  ft. 
from  the  nearest  point  of  a  sphere  20  ft.  in  diam- 
eter. Find  the  area  of  that  portion  of  the  sphere 
which  is  illuminated  by  the  light  at  A. 

Ex.  94.  How  far  is  the  light  A  from  the  surface  when  the  illuminated 
portion  of  the  sphere  is  ^  of  the  whole  surface  ?  When  the  illuminated 
portion  is  ^  of  the  whole  ? 

Ex.  95.  Why  is  the  illuminated  portion  of  the 
surface  always  less  than  |  the  whole  surface  ? 

Ex.  96.  A-BGE  is  a  zone  of  altitude  AC.  If 
AB,  BF  be  drawn,  what  is  the  value  of  Z  ^  ? 

In  the  A  ABF^  what  relation  connects  AB,  AC, 
AF?  (XVII.  5  (a).) 

Prove  that  the  area  of  the  zone  A-BGE  =  the 
area  of  a  circle  of  which  AB  is  the  radius. 

Ex.  97.  The  altitude  CA  of  the  zone  A-BFE 
equals  H)  the  radius  of  the  sphere  equals  B. 

What  is  the  area  of  the  zone  ?  What  is  the  vol- 
ume of  the  spherical  sector  S-ABFF?  Show  that, 
from  the  rt.  A BCS,  BC^  =  2E-H-H\ 

Find  the  expression  for  volume  of  cone  S-BFE. 

Ex.  98.  Show  that  the  volume  of  the  spherical 
segment 


A~BFE  = 


2  ttB'-H     t(B 


H)(2B'  H-m) 
3 


~1LKs2B^ 
3    ^ 


(iB-II)(2B 


^)} 


=  ^'(3i?-^). 


XXVI.    GROUP  ON  GEOMETRY  OF  THE  SPHERE 
SURFACE;   BRIEFLY,  SPHERICAL  GEOMETRY 

DEFINITIONS 

The  Distance  between  two  points  on  the  surface  is  the  shorter 
arc  of  a  great  circle  that  passes  through  the  points. 

The  Arc-Radius  of  a  small  circle  on  the  sphere  is  the  shorter 
polar  distance  of  any  one  of  its  points. 

A  Spherical  Angle  is  the  figure  formed  by  two  great-circle 
arcs  that  intersect  in  one  point. 

The  Angle  between  any  two  circles  that  have  a  common  point  is 
the  plane  angle  formed  by  tangents  to  the  circles  at  this  point. 

A  spherical  angb,  therefore,  is  the 
same  as  the  angle  between  the  tangents 
to  its  sides  drawn  at  the  point  of  inter- 
section. 

Thus,  the  spherical  angle  ABC  is  the 
same  as  the  plane  angle  FBH  formed  by 
the  tangents  FB  and  HB,  to  the  sides  AB 
and  BC,  respectively,  at  B.  ' 

Corollaries  to  the  Definitiox 

(a)  Any  spherical  angle  is  the  measure  of  the  dihedral 
formed  by  the  -planes  of  its  sides. 

(b)  The  measure  of  a  spherical  angle  (ABC),  plane 
angle  FBH,  is  the  are  AC  intercepted  hy  its  sides  on  the 
great  circle  of  whi<ih  its  vertex  (B)  is  the  pole. 

Dem.     Draw  AS,  BS,  C^  to  the  center  S. 

Z  AjSB  =  Z  CSB  =  rt.  Z.         (Def .  of  pole.) 
Z  FBS  =  Z  HBS  =  rt.  Z.  (IX,  4.) 

328 


XXVI.     SPHERICAL   GEOMETRY  329 

.-.  FB  II  AS  and  HB  II  CS.  (Def.  of  lis.) 

.-.  Z  FBH=  Z  ASC.  (XXI.  7.) 

But  angle  ASC  is  measured  by  arc  AC.  (XII.  1.) 

.'.  Z.FBHis  measured  by  are  AC. 

Q.E.D. 

A  Spherical  Polygon  is  a  portion  of  the 
surface  of  the  sphere  bounded  by  arcs  of 
great  circles. 

A  Convex  Polygon  is  one  the  perimeter 
of  which  caunot  be  cut  by  a  great  circle  in 
more  than  two  points.  Only  convex  poly- 
gons will  be  discussed  in  this  group. 

Spherical  polygons  are  classified  in  the 
some  manner  as  plane  polygons. 

A  right  spherical  triangle  is  said  to  be  Rectangular,  when  it 
contains  one  right  angle ;  Birectangular,  when  it  has  two  right 
angles,  and  Trirectangular,  when  it  contains  three  right  angles. 

The  term  right  triangle  will  hereafter  be  used  to  denote  a 
spherical  triangle  with  hut  one  right  angle,  unless  the  context 
evidently  implies  the  existence  of  more  than  one  such  angle  in 
the  triangle. 

Corollaries  to  the  Definition 

(a)  A  polyhedral  that  has  Us  vertex  at  the  center  of 
the  sphere  cuts  from  the  surface  a  spherical  polygon 
whose  sides  are  the  measures  of  the  face  angles  of  the 
polyhedral  and  whose  angles  are  the  measures  of  the 
dihedrals  of  the  polyhedrals. 

(h)  The  sum  of  any  two  sides  of  a  spherical  triangle 
is  greater  than  the  third.  (XXII.  (h)  3.) 

(c)  The  sum  of  the  sides  of  a  spherical  polygon  {i.e.  the 
perimeter)  is  less  than  the  circumference  of  a  great 
circle.  (XXII.  QS)  4.) 

(d)  An  isosceles  spherical  triangle  is  isoangular. 

(XXII.  Ex.  29.> 


880         THE  ELEMENTS  OF  GEOMETRY 

Two  spherical  polygons  are  Symmetric, 
when  the  parts  of  the  one  are  respec- 
tively equal  to  the  parts  of  the  other, 
but  are  arranged  in  opposite  order. 

If  two  symmetric  polyhedrals  have 
their  vertex  at  the  center  of  the  sphere, 
they  cut  from  the  surface  two  symmetric 
spherical  polygons. 

Thus  Z  AOB  =  A  A' OB'.  (Vertical  angles.) 

.*.  arc  AB  =  arc  A'B'.     (Measures  of  equal  angles.) 

Again,      dihedral  A-BO-C  =  dihedral  A'-B'0-C\ 

(YeTtical  dihedrals.) 
.-.  AB—  /LB\     (Measures  of  =  dihedrals.) 

Similarly,  for  the  other  parts  of  the  two  figures. 

The  opposite  order  of  parts  in  the  two  figures  is  shown  by 
the  arrow  heads. 

Symmetric  polygons  so  situated  that  the  joins  of  correspond- 
ing vertices  concur  at  the  center  of  the  sphere  are  said  to  be  in 
Perspective  with  regard  to  this  center. 

Two  symmetric  polygons  cannot,  in  general,  be  placed  in 
coincident  superposition. 

For,  on  account  of  the  curvature  of  the  sphere  surface,  it  will 
be  impossible  to  make  the  parts  of  the  one  coincide  with  the 
corresponding  parts  of  the  other  without  breaking  or  tearing 
the  surface. 

For  the  same  reason,  it  is  impossible  to  revolve  one  of  two 
symmetric  polygons  having  homologous  sides  in  common  about 
this  common  side,  so  as  to  make  the  polygons  coincide.  That 
is,  revolution,  as  a  step  in  a  proof,  cannot  occur  in  the  geometry 
of  the  sphere-surface ;  an  arc  cannot  be  taken  as  an  axis. 

It  will  be  shown,  however  (XXVI.  3  a),  that  if  polygons  are 
symmetric,  they  are  equal  in  area. 

A  spherical  polygon  is  the  Polar  of  a  second,  when  the  ver- 
tices of  the  first  are  the  poles  of  the  sides  of  the  second. 

A  spherical  polygon  is  Supplemental  to  a  second,  when  the 


XXVI.     SPHERICAL   GEOMETRY  331 

angles  of  the  first  are  the  supplements  of  the  corresponding 
sides  of  the  second. 

The  Spherical  Excess  of  a  polygon  of  %  sides  is  the  excess  of 
the  sum  of  its  angles  over  (2  n  —  4)  right  angles. 

A  Lune  is  a  portion  of  a  sphere-surface  bounded  by  two  great 
semicircles. 

An  Ungula  or  Spherical  Wedge  is  the  portion  of  the  volume  of 
the  sphere  comprised  between  two  great  Semicircles. 

Corollaries  to  the  Definitions. 

(a)  A  lune  is  the  same  fraction  of  the  total  surface  of 
a  sphere  that  the  angle  of  the  lune  is  of  four  right 
angles. 

(b)  An  ungula  is  the  same  fraction  of  the  volujne  of 
the  sphere  that  the  angle  of  the  ungula  is  of  four  right 
angles. 

Preliminary  Scholium 

Correspondence  between  Plane  and  Spherical  Geometry 

The  Line.  Since  any  two  points  (not  the  extremities  of  a 
diameter)  determine  a  great  circle  of  the  sphere  (XXV.  1  g), 
the  great-circle  of  the  sphere  corresponds  to  the  straight  line  of 
the  plane. 

A  great-circle  arc  may  therefore  be  called  simply  a  line  of  the 
sphere-surface. 

But  two  great-circle  arcs  perpendicular  to  a  third  meet  in 
the  poles  of  this  third,  and  from  either  of  these  poles  an  infinite 
number  of  perpendiculars  can  be  drawn  to  the  great-circle  arc. 

Parallels.  Hence,  there  are  no  lines  on  the  sphere  corre- 
sponding to  the  parallels  of  the  plane. 

Axioms.     It  follows  that : 

(1)  There  is  no  parallel  axiom  on  the  sphere  surface. 

(2)  All  the  axioms  of  the  plane,  except  Axioms  7  and  8,  are 
true  on  the  surface  of  the  sphere. 


882         THE  ELEMENTS  OF  GEOMETRY 

The  Circle.  Three  non-coUinear  points  in  a  plane  determine 
a  circle. 

Three  non-collinear  points  on  a  sphere-surface  determine  a 
small  circle  of  this  surface. 

The  small  circle  of  the  sphere- surface  therefore  corresponds  to 
the  circle  of  the  plane,  the  word  Pule  (meaning  the  nearer  pole) 
replacing  the  word  Center  of  the  plane  geometry. 

Propositions.  The  propositions  of  spherical  geometry  are 
therefore  derived  from  those  of  plane  geometry  by  replacing 
the  "  straight  line  "  of  the  plane  by  the  "  great-circle  arc,"  or 
simply  the  "line"  of  the  sphere-surface,  and  dropping  all 
propositions  that  cannot  he  proved  without  using  Axiom  7  or 
Axiom  8. 

The  proofs  for  the  sphere-surface  are  identical  with  those 
already  given  for  the  plane  where  Axioms  7  and  8  are  not 
involved. 

Summary  of  Propositions  Common  to  Plane  and  Spherical 
Geometry 

The  following  summary  presents  the  geometry  of  the  sphere 

as  it  corresponds  to  the  geometry  of  the  plane. 

Plane  Sphekb 

Group   I.  All  propositions  true  on  the  sphere. 

Group   II.  No  propositions  true  on  the  sphere. 

Group   III.  No  propositions  true  on  the  sphere. 

Group   IV.  All  propositions  true  on  the  sphere. 

Group   V.  All  propositions  true  on  the  sphere, 

if  "congruent"  be  replaced  by 
"  congruent  or  symmetric." 

Group   VI.  No  propositions  true  on  the  sphere. 

Group  VII.  Only  Theorems  1  and  2  true  on  the 

sphere. 

Groups  VIII.,  IX.,  X.,  XI.     All  propositions  true  on  the  sphere. 

Group   XII.  Only  Theorem  1  true  on  the  sphere, 

but  not  used  as  a  basis  of  meas- 
urement. 


XXVI.     SPHERICAL   GEOMETRY 

Areal  Measurement  in  the  Plane.  The  plane  is  boundless  (i.e. 
without  boundaries  of  any  sort),  and  is  infinite  in  extent.  The 
ratio  of  the  surface  of  a  plane  figure  to  the  total  surface  of  the 
plane  on  which  it  lies  cannot  be  expressed  in  numbers.  Hence, 
a  purely  arbitrary  unit,  the  square  on  the  linear  unit,  is  selected, 
and  the  areal  unit  changes  when  the  linear  unit  changes.  All 
the  propositions  of  plane  geometry  that  deal  with  area  involve 
the  use  of  the  square, 

Neither  the  square  nor  any  analogous  figure  can  be  drawn  on 
the  sphere-surface. 

Areal  Measurement  on  the  Sphere.  The  sphere-surface,  though 
boundless,  is  not  infinite  (XXV.  4).  The  ratio  of  the  surface 
of  a  spherical  polygon  to  the  whole  sphere-surface  on  which  it 
lies  can  be  expressed  in  numbers.  Hence,  the  sphere-surface 
itself  (or  some  definite  fraction  of  it)  is  the  natural  unit  of  area 
for  the  figures  that  lie  upon  it,  a  unit  that  never  changes  for  a 
given  surface. 

Accordingly,  the  treatment  of  surface  measurement  on  the 
sphere  is  essentially  different  from  the  treatment  of  this  subject 
in  the  plane,  and  no  propositions  of  plane  geometry  that  deal 
with  areas  or  areal  relations  can  have  any  correspondents  on 
the  sphere-surface. 

Again,  the  Doctrine  of  Similarity,  with  all  its  varied  applica^ 
tions,  is  based  on  Axioms  7  and  8. 

There  is,  therefore,  no  Theory  of  Similarity  in  spherical 
geometry. 

For  these  reasons,  no  use  can  be  made  of  the  plane  geometry 
beyond  Group  XII. 

Detailed  proofs  will  be  required  only  for  those  propositions 
which  are  peculiar  to  the  sphere-surface,  or  which  have 
been  proved  in  plane  geometry  by  use  of  Axioms  7  and  8, 
when  a  proof  not  involving  these  axioms  might  have  been 
used. 


884  THE  ELEMENTS  OF  GEOMETRY 

PROPOSITIONS 

XXVI.  1.  The  shortest  line  that  can  he  drawn  on  the 
surface  of  the  sphere,  between  any  two  points  c^i  the 
surface,  is  the  great-circle  arc  {not  greater  than  a  semi- 
circumference)  that  joins  them. 


Hyp.  If  AB  is  a  great-circle  arc  less  than  a  semicircle,  and 
ALFB  any  other  line  whatever  from  Ato  B, 

Cone:  then  AB  <  ALFB. 

Dem.  With  ^  as  a  pole  and  arc-radius  less  than  ABy  de- 
scribe O  My  cutting  AB  in  G  and  ALFB  in  L. 

With  5  as  a  pole  and  an  arc-radius  BC,  describe  O  Q,  cutting 
ALFB  in  F,     This  circle  will  be  tangent  to  O  iWat  G. 

(IX.  8  a,  converse  (a).) 

Whatever  the  form  of  the  line  AL  may  be,  a  line  equal  to 
AL  may  be  drawn  from  A  to  G. 

For,  if  the  zone  whose  base  is  O  Jf  be  revolved  on  the  axis 
of  O  Jf,  Ly  moving  along  the  circumference,  will  reach  the  posi- 
tion C,  and  the  line  AL  will  itself  extend  from  A  to  G. 

Similarly,  a  line  equal  to  BF  can  be  drawn  from  B  to  G. 

But  L  is  without  O  Q  (O  Q  is  externally  tangent  to  O  Jf  at  (7). 
.-.  BF+  FL  >  BG.  (Preliminary  Th.  1.) 

.-.  BF  +  FL  +  LA>  BG  +  GA]     (Preliminary  Th.  1.) 
t.e.  BFLA  >  BA. 

Q.E.D. 


XXVI.     SPHERICAL   GEOMETRY  336 

ScH.  If  A  and  B  be  extremities  of  a  diameter,  an  infinite 
number  of  equal  great-circle  arcs  may  be  drawn  from  A  to  B, 
but  each  will  be  less  than  any  line  not  a  great-circle  arc  that 
can  be  drawn  from  A  to  B. 

Note.  —  In  speaking  of  two  points  on  the  sphere,  we  shall  hereafter 
assume,  unless  the  contrary  be  stated  or  evidently  implied,  that  these 
points  are  not  the  extremities  of  a  diameter  of  the  sphere. 

The  proof  of  the  theorem  emphasizes  the  analogy  between  the  great 
circle  on  the  sphere  and  the  straight  line  of  the  plane.  Each  is  deter- 
mined by  two  points  of  the  surface  ;  each  measures  the  shortest  distance 
on  the  surface  between  any  two  of  its  points. 

Ex.  1.   Show  that  a  spherical  triangle  may  have  3  obtuse  angles. 

Ex.  2.  Two  spherical  polygons  of  the  same  number  of  sides  are  equal, 
if  the  sum  of  the  angles  of  the  one  equals  the  sum  of  the  angles  of  the 
other. 

Ex.  3.  Two  spherical  right  triangles  are  congruent  or  symmetric,  if  the 
oblique  angles  of  the  one  equal  respectively  the  oblique  angles  of  the  other. 

Ex.  4.  Two  birectangular  triangles  are  congruent  or  symmetric,  if  the 
oblique  angle  of  the  one  equals  the  oblique  angle  of  the  other. 

Ex.  5.  If  the  diagonals  of  a  spherical  polygon  bisect  each  other,  the 
opposite  sides  are  equal. 

Ex.  6.  If  two  spheres  intersect,  the  tangent  lines  drawn  to  the  spheres 
from  any  point  in  the  plane  of  the  circle  of  intersection  are  equal. 

Def.  If  two  spheres  be  generated  by  the  revolution  of  two  circles  about 
their  line  of  centers  as  an  axis,  the  radical  axis  of  the  circles  (XVII.  Ex. 
72,  Def. )  will  generate  a  plane  perpendicular  to  the  line  of  centers  of  the 
spheres.  (XXI.  3.) 

This  plane  is  called  the  Radical  Plane  of  the  Spheres. 

Ex.  7.  The  radical  plane  of  two  spheres  is  the  locus  of  the  point  from 
which  pairs  of  equal  tangent  lines  may  be  drawn  to  the  spheres. 

Ex.  8.  The  radical  planes  of  three  spheres  taken  two  and  two  inter- 
sect in  a  line  perpendicular  to  the  plane  of  the  centers  of  the  spheres. 

The  radius  of  a  sphere  is  20  ft.    Find  the  area  : 

Ex.  9.  Of  a  triangle  on  the  sphere  whose  angles  are  82°,  104°,  and  84°, 
respectively. 

Ex.  10.    Of  a  birectangular  triangle  whose  vertex  angle  is  72°. 

Ex.  11.   Of  a  trirectangular  triangle. 


886  THE   ELEMENTS  OF  GEOMETRY 

XXVI.  2.  Isosceles  symmetric  triangles  are  congruent. 


Hyp.     If  the  A  ABC,  A'B'C  are  symmetric,  and  if  AB 
=:AC,A'B'  =  A'a, 

Cone:  then  A  ABC  ^  A  A'B'C. 

Dem.  AB  =  AC=A'B'=A'C'.  (Hyp.) 

Place  A  ABC  on  A  A'B'C,  so  that  A  shall  fall  at  A'  and  B 
shall  fall  at  C. 

The  two  triangles  fall  on  the  same  side  of  A'C,  since  the  parts 
occur  in  reverse  order. 

ZA  =  Z  A'.  (Def.  Sph.  n-gon.  d.) 

.'.  AC  takes  the  direction  oi  A'B'. 

AC=A'B'.  (Hyp.) 

/.  C  falls  at  B'. 

.'.  BC  coincides  with  B'C 

[Two  points,  etc.,  determine  a  great  circle.]         (XXV.  1  g.) 

.-.  A^^C^A^'^'C". 

Q.E.D. 


XXVI.     SPHERICAL   GEOMETRY  837 

XXVI.  3.  Symmetric  triangles  are  equal. 


Hyp.     If  A  ABC  is  symmetric  to  A  A'B'C, 
Cone:  then  AABC  =  AA'B'C'. 

Dem.  Place  A  ABC  in  perspective  with  AA'B'C,  with 
respect  to  S.  Let  Q  be  the  pole  of  the  circle  through  A,  B, 
and  C. 

[Three  points,  etc.,  determine  a  small  circle.]       (XXV.  1  /.) 

Draw  QSQ'  and  the  joins  QA',  Q'B',  and  Q'C. 

Trihedral  S-ABQ  is  symmetric  with  trihedral  S-A'B'Q'. 

(Def.  of  symmetric  trihedrals.) 

.-.  AABQis  symmetric  with  AA'B'Q'. 

(Def.  of  symmetric  w-gons,  etc.) 

.-.  QA=  Q'A'  and  QB=  Q'B'. 

But  QA  =QB=  Q'A'  =  Q'B' ;  (XXV.  1  e.) 

i.e.  A  AQB  and  A' Q'B'  are  isosceles. 

.:  AAQB^A  A'Q'B\  (XXVI.  2.) 

Similarly,  A  BQC  ^  A  B'Q'O, 

and  AAQG^AA'Q'Q'. 

.\AAQB+ABQG-{-AAQC=AA'Q'B'-^AB'Q'C'-hAA'Q'a; 

K&  A  ABC  =  AA'B'C. 

Q.E.D. 


838         THE  ELEMENTS  OF  GEOMETRY 

XXVI.  3  a.  Symmetric  polygons  are  equal 

Outline  Dem.    Divide  the  polygons  into  triangles  by  diagonals 

from  homologous  vertices. 

These  triangles  will  be   symmetric  in  pairs  and  therefore 

equal  in  pairs,  therefore  the  polygons  are  equal. 

Q.E.D. 

XXVI.  4.    If  one  triangle  is  polar  to  another,  the 
second  is  polar  to  the  first. 


Hyp.    If  A,  B,  C,  are  Llie  poles  of  B'C,  A'C,  A'B\  respectively, 

Cone.  :  then  A\  B\  C",  are  the  poles  of  BC,  AC,  AB,  respec- 
tively. 

Dem.  A  is  the  pole  of  B'C,  (Hyp.) 

.-.  ^(7'  =  90°. 

[The  arc-radius  of  a  great  O  is  a  quadrant  (XXV.  1  e.)  Sch.] 

Similarl}^,  BC  =  90°. 

.*.  A  and  B  lie  in  a  great  circle  of  which  C  is  the  pole. 
But  only  one  great  O  can  pass  through  A  and  B,  (XXV.  Ig.) 
and  the  side  AB  is  a  great-circle  arc. 

.-.  C  is  the  pole  of  the  side  AB. 


XXVI.     SPHERICAL   GEOMETRY  339 

Similarly,  B'  is  the  pole  of  the  side  AC  and  A  is  the  pole  of 

the  side  BG.  ^  „  ^ 

Q.E.D. 


ScH.     If  the  sides  of  AA'B'C  be  ex- 
tended to  meet  in  A",  B",  C",  three  new 
triangles  are  formed  each  of  which  is  B\ 
polar  to  A,  B,  O.     A  triangle  therefore 
has  four  polars. 


But  in  speaking  of  the  polar  of  a  triangle,  we  always  mean 
the  central  figure,  for  which  each  of  the  distances  Aa\  Bb', 
Co',  is  less  than  a  quadrant. 

XXVI.  4  a.  Jf  one  polygon  is  polar  to  another,  the 
second  is  polar  to  the  first. 

XXVI.  4:b.  A  trirectangular  triangle  is  its  own  polar 
(i.e.  is  self-polar). 

The  radius  of  a  sphere  is  20  feet.     Find  the  area  : 

Ex.  12.  Of  a  quadrilateral  whose  angles  are  112°,  85°,  92°,  and  126°, 
respectively. 

Ex.  13.  Of  a  pentagon,  whose  angles  expressed  in  radians,  are  2.25, 
2.83,  2.7,  3.72,  and  1.7,  respectively. 

Ex.  14.    Of  a  hexagon  each  of  whose  angles  is  130°. 

The  radius  of  a  sphere  is  20  ft.  A  plane  passes  through  the  sphere  at 
a  distance  12  ft.  from  the  center.     Find  : 

Ex.  15.  The  total  surface  of  the  minor  spherical  segment  cut  off  by  the 
plane. 

Ex.  16.  The  angles  of  an  equiangular  triangle  on  the  sphere  whose 
area  is  equal  to  the  zone  surface  of  the  segment. 

Ex.  17.  The  angles  of  an  equiangular  triangle  equal  in  area  to  the  total 
surface  of  the  segment. 

Ex.  18.  The  third  angle  of  a  triangle,  two  of  whose  angles  are  100° 
and  75°,  and  which  equals  in  area  one  face  of  the  inscribed  cube. 


840  THE  ELEMENTS  OF  GEOMETRY 

XXVI.  6.  Polar  triangles  are  supplemental, 

0' 

Hyp.  If 
A  ABC  is 
the  polar  of 

Cone:  thenZ^=180"'-a';  ZB=180°-6';  Za=180°-c', 
and  Z^'  =  180°-a;   Z5'=180°-6;  ZC'  =  180°-c. 

Dem.  Extend  AB,  AC  to  meet  B'O  as  at  E  and  F.  Z.A\3 
measured  by  FE.  (Def.  of  measure  of  spherical  Z.) 

C  is  the  pole  of  ABE.  (Def.  of  polar  n-gons.) 

.-.  C'E  =  90°', 

i.e.  C'F+FE  =  90°. 

Similarly,  •  FB'  =  90°. 

.-.  C'F-i-FB' -{-FE  =  1S0°', 

i.e,  C'B'  +  FE  =  180°. 

.-.  FE  =  180°  -  C'B', 

or  Z^  =  180°-a'. 

Similarly,  for  Z  B  and  Z  (7. 

In  like  manner,  we  may  show  that 

Z.A'  =  180°  —  a  (by  extending  BC  as  shown),  etc. 

Q.E.D. 

XXVI.  5  a.  Polar  polygons  are  supplemental. 

XXVI.  5  h.  Two  triangles  are  congruent  or  symmetinc, 
if  the  angles  of  the  one  are  equal  respectively  to  the 
angles  of  the  other. 

Dem.     Let  P  and  P  be  the  polars  of  the  given  triangles. 
The  sides  of  P  equal  the  sides  of  P'.  (XXVI.  5.) 

.-.  P  is  congruent  (or  symmetric)  to  P.  (V.  3.) 


XXVI.     SPHERICAL   GEOMETRY  341 

.*.  the  angles  of  P  equal  the  angles  of  P. 

•  (Def.  of  symmetric  n-gons.) 

.*.  the  sides  of  the  given  triangles  are  equal,  respectively. 

(XXVI.  5.) 

.*.  the  given  triangles  are  congruent  or  symmetric.        (V.  3.) 

Q.E.D. 

XXVI.  6.    The  sum  of  the  interior  angles  of  a  tri- 
angle lies  hetiveen  two  right  angles  and  six  right  angles. 


Hyp.     If  ABC  is  a  spherical  triangle, 
Cone:  thenZ^  +  ZS  +  ^C>2  rt.z^and<6  rt.  A 
Dem.     Let  A  ^'^'C  be  the  polar  of  A  ABC. 
Then  Z  J  =  2  rt  A  -  a'.  (XXVI.  5.) 

Z  7i  =  2  rt.  A  -  b'. 
Z(7=2rt.Zs-c'. 
.-.  Zyl  +  Z5  +  Z<7=6rt.Zs-(a'-f  6'4-c'). 
But  a'  +  ft'  4-  c'  >  0  and  <  4  rt.  A.  (Cor.  to  the  def.  XXVI.  c.) 
.-.  Z  ^  +  Z5  -}-  Z  (7>  6  rt.  Z  -  4  rt.  Zs,  or  2  rt.  A, 

and  <  6  rt.  Z  —  0,  or  6  rt.  Z. 

Q.E.D. 

XXVI.  6  a.  The  sum  of  the  interior  angles  of  a  jjohj- 
gon  lies  hetiveen  (2  n  —  4)  right  angles  and  2  n  right 
angles  (n  being  the  numher  of  sides  of  the  polygon). 


842  THE   ELEMENTS   OF  GEOMETRY 

XXVI.  7.  If  two  great  semicircles  intersect  on  the 

surface  of  a  hemisphere,   the   sum   of  either  set  of 

opposite  triangles  equals  the   lune  whose  angle  is  the 
corresponding  angle  of  the  semicircles. 


Hjrp.     If  the  lines  ACA',  BCB'  meet  the  base  of  the  hemi- 
sphere S  in  Af  A\  By  B', 

Cone. :  then  A  ABO  +  A  A'B'C  =  lune  OA'C'B'. 

Dem.     CA'  +  A'C'  =  CA'C  =  180°.  (Def .  of  great  circle.) 

CA'  +  AC  =  ACA'  =  180°.  (Same  reason.) 

.-.  CA'  +  AC=  CA'  +  A'C.  (Ax.  1.) 

.-.  AC=A'a.  (Ax.  2.) 

Similarly,  BC=B'C', 

and  AB  =  A'B'. 

.'.  A  ABC  is  symmetric  and  equal  to  A  A'B'C.     (XXVI.  3.) 

.-.  A  ABC  +  A  A'B'C  =  A  A'B'C  +  A  A'B'C    (Ax.  2.) 

=  lune  CA'CB'  (or  lune  C). 

Q.ED. 


XXVI.     SPHERICAL   GEOMETRY  348 

XXVI.  7  a.  A  trirectangular  triangle  is  one  eighth 
of  the  surface  of  the  sphere  on  which  it  lies. 

ScH.  Unit  of  Area.  Angular  Unit.  The  trirectangular 
triangle,  or  one  eighth  of  the  sphere-surface,  will  be  taken  as 
the  unit  of  areal  measure  on  the  sphere,  and  the  right  angle  as 
the  unit  of  angle  measure.  Angles  must  be  expressed  in  terms 
of  the  right  angle  in  all  formulae  for  area. 

Thus,  a  lune  whose  angle  is  a  right  angle  is  twice  the  tri- 
lectangular  triangle;  i.e.  its  area  is  2;  a  lune  whose  angle  is 
50°,  or  I  rt.  Z,  is  f  of  2,  or  -LQ  ;  a  lune  whose  angle  is  A  has  an 
area  of  2  A,  etc.  The  area  of  the  sphere  is  8;  that  of  the 
hemisphere  is  4. 

XXVI.  8.  The  area  of  a  spherical  triangle  equals  its 
spherical  excess.  ^ 


Hyp,     If  the  angles  of  the  A  ABC  be 

expressed  in  rt.  A,  and  the  trirectangular 
A  be  taken  as  the  unit  of  area, 


Cone. :  then  the  area  of  A  ABC  =^  +  ^  +  (7—2. 

Dem.     On  the  hemisphere  ABB'A\ 

A  ABC  -{-AAB'C  =  lune  B  =  2B,     (XXVI.  7.  Sch.) 

A  ABC  +  A  A'BC  =  lune  A  =  2A,     (Same  Eeason.) 

A  ABC  -\-AA'B'0  =  lune  (7=2  0.     (Same  Eeason.) 
But  ^ 

A  ABC+A  AB'G+A  A'BC+A  A'B'C^the  hemisphere 

=  4.        (XXVI.  7.  Sch.) 

.-.  (by  addition)  2  A  ABC  +  4  =  2  (^  +  J5  +  C), 

or  AABC  +  2  =  A-\-B+C. 

.:  AABC  =  A-\-B+C-2. 

Q.E.D. 


344  THE  ELEMENTS  OF  GEOMKTRY 

XXVI.  8  a.  The  area  of  a  spherical  polygon  equals 
its  spherical  excess. 

Let  the  student  supply  the  proof  by  dividing  the  polygon 
into  triangles  and  using  the  theorem. 

Ex.  19.  The  angles  of  an  equiangular  triangle  equal  in  area  to  the  con- 
vex surface  of  the  cone  of  tangents  extending  to  the  sphere  from  a  point 
whose  distance  from  the  center  is  35  ft. 

Note. — The  few  exercises  appended  to  Group  XXVI  are  intended 
merely  as  illustrations  of  the  method  of  utilizing  the  theorems  and  exer^ 
cises  of  the  plane  geometry  as  exercises  in  spherical  geometry.  As  stated 
in  the  General  Scholium,  every  proposition  of  the  plane  geometry  that 
does  not  involve  Ax.  7  or  Ax.  8  as  a  necessary  element  in  its  proof,  is 
true  on  the  sphere-surface  as  in  the  plane. 

In  the  following  problems,  the  word  line  (unqualified)  means  arc  of  a 
great  circle;  the  word  circle  and  the  symbol  O  (unqualified)  mean  small 
circle.     All  data  are  supposed  to  be  given  on  the  sphere-surface. 

What  is  the  locus  of  a  point : 

Ex.  20.  At  a  given  distance  from  a  given  point  ? 

Ex.  21.  At  a  given  distance  from  a  given  line  ? 

Ex.  22.  At  a  given  distance  from  a  given  circle  of  radius  r  ? 

Ex.  23.  Equidistant  from  two  given  points  ? 

Ex.  24.  Equidistant  from  two  given  lines  ? 
What  is  the  locus  of  the  centers  of  circles  : 

Ex.  25.  Of  given  radius,  passing  through  a  given  point  ? 

Ex.  26.  Of  given  radius  tangent  to  a  given  line  ? 

Ex.  27.  Of  given  radius  tangent  to  a  given  circle  ? 

Ex.  28.  Passing  through  two  given  points  ? 

Ex.  29.  Tangent  to  two  given  lines  ? 

Ex.  30.  Tangent  to  two  concentric  circles  ? 

Ex.  31.  Tangent  to  two  equal  circles? 

Ex.  32.  Bisect  a  given  line-segment. 

Ex.  33.  Circumscribe  a  O  about  a  A. 

Ex.  34.  Bisect  a  given  Z. 

Ex.  35.  Inscribe  a  O  in  a  given  A. 

Ex.  36.  Escribe  a  O  to  a  given  ^ 


XXVI.     SPHERICAL   GEOMETRY  345 

XXVI.     SUMMARY   OF    PROPOSITIONS    IN    THE   GROUP 
ON   SPHERICAL   GEOMETRY 

1.  The  shortest  line  that  can  he  draivn  on  the  sj^here, 
between  any  two  points  on  the  surface,  is  the  great-circle 
arc  (not  greater  than  a  semicircumference)  that  joins 
them. 

ScH.  There  is  no  "  shortest  line  "  on  the  sphere-surface  be- 
tween the  extremities  of  a  diameter  of  the  sphere. 

2.  Isosceles  symmetric  triangles  are  congruent. 

3.  Symmetric  triangles  are  equal. 

a.  Symmetric  polygons  are  equal. 

4.  If  one  triangle  is  polar  to  another,  the  second  is 
polar  to  the  first. 

ScH.     A  triangle  has  four  polars. 

a.  If  one  polygon  is  polar  to  another,  the  second 

is  polar  to  the  first, 
h.  A    trirectangular    triangle   is    its    own  polar 

(i.e.  is  self-polar), 

5.  Polar  triangles  are  supplemental. 

a.  Polar  polygons  are  supplemeiital. 

b.  Tivo  triangles  are  congruent  or  symmetric,  if 

the  angles  of  the  one  are  equal  respectively  to 
the  angles  of  the  other. 

6.  The  sum  of  the  interior  angles  of  a  triangle  lies 
hetioeen  tivo  right  angles  and  six  right  angles. 


346         THE  ELEMENTS  OF  GEOMETRY 

a.  The  sum  of  the  intenor  angles  of  a  polygon  lies 
heiiveen  (2ii-4)  right  angles  and  2n  right 
angles y  n  being  the  number  of  sides  of  the 
polygon. 

7.  If  two  great  semicircles  intersect  on  the  surface 
of  a  hemisphere,  the  sum  of  the  opposite  triangles 
formed  equals  the  lune  whose  angle  is  the  angle  of 
the  two  semicircles. 

a.  A  trirectangular  triangle  is  one  eighth  of  the 
surface  of  the  sphere  on  which  it  lies. 

ScH.    Unit  of  Area  and  Angular  Unit. 

8.  The  area  of  a  spherical  triangle  equals  its  spheri- 
cal excess. 

a.  The  area  of  any  spherical  polygon  equals  its 
spherical  excess. 


NOTES  AND  BIOGRAPHICAL  SKETCHES 

Alexandrian  School  (The  First).  From  earliest  times  to  about  100  n.c. 
To  tliis  school  belonged  Euclid.  The  Second  Alexandrian  School  began 
with  the  Christian  Era.  Menelaus,  Theon,  and  Hypatia  belonged  to  this 
school. 

ApoUonius,  Greek.  Lived  in  third  century  b.c.  He  ranked  with 
Euclid  and  Archimedes.     Known  as  the  "Great  Geometer." 

Archimedes,  Greek.  Third  century  b.c.  Born  in  Syracuse.  Greatest 
mathematician  of  antiquity. 

Ceulen,  Ludolp  van,  Netherlands.  Carried  the  value  of  ir  to  36  places, 
known  as  Ludolp's  Numbers. 

Ceva,  G.  (1048-1737),  Italian.  Discovered  the  theorem  that  bears  his 
name. 

Chasles,  M.  (1793-1880),  French.  He  did  much  to  develop  modern 
projective  geomet^-y. 

Cissoid.  A  curve  by  means  of  which  may  be  found  two  mean  propor- 
tionals between  two  given  straight  lines. 

Cube,  Duplication  of.  One  of  the  three  ancient,  unsolved  problems  of 
geometry,  the  other  being  the  quadrature  of  a  circle  and  the  trisection  of 
an  angle,  to  which  the  ancients  devoted  much  time.  It  is  now  generally 
admitted  that  they  cannot  be  solved  by  the  geometry  of  the  straight  line 
and  circle. 

Descartes,  RenI  (1596-1650),  French.  He  introduced  into  geometry 
the  use  of  the  algebraic  equation  for  the  purpose  of  analysis. 

Euclid  (about  300  B.C.),  Greek.  Lived  in  Alexandria.  Probably  went 
there  from  Athens.  Not  much  is  reliably  known  of  him.  His  "Elements," 
containing  13  books,  has  been  the  basis  of  all  elementary  geometrical 
instruction  for  over  2000  years.  His  geometry  does  not  touch  upon  the 
subject  of  mensuration. 

Eudoxus  (408  b.c).  Pupil  of  Plato.  Astronomer  and  legislator,  as  well 
as  mathematician. 

Euler,  L.  (1707-1783),  Swiss.  He  wrote  a  great  number  of  works  on 
mathematics.  He  introduced  much  of  current  notation  into  trigononaetry  ; 
also  use  of  tt. 

Golden  Section.  It  cuts  a  line  in  extreme  and  mean  ratio.  It  was  much 
studied  by  Eudoxus. 

347 


348  THE   ELEMENTS   OF   GEOMETRY 

Harmonics.  Its  fundamental  theorem  was  discovered  by  Serenus  of  tlie 
Second  Alexandrian  School. 

Hippocrates  of  Chios  (4:50  n.c),  Greek.  Developed  the  subject  of 
similar  figures ;  also  discovered  important  relations  between  areas  of 
circles.     He  was  the  first  author  of  an  elementary  text-book  on  geometry. 

Legendre,  A.  M.  (1762-18;i3),  French.  Wrote  an  "  Elements  of  Geome- 
try "  in  1794,  which  was  largely  adopted  in  continental  Europe  and  the 
United  States  as  a  substitute  for  the  more  difficult  Euclid. 

Menelaus  (98  a.d.),  Greek.  Of  the  Second  Alexandrian  School.  The 
theorem  of  Menelaus  is  the  foundation  of  the  modern  theory  of  trans- 
versals.    He  contributed  largely  to  our  knowledge  of  trigonometry. 

IT.     For  its  history  and  values  see  Cajori's  "  History  of  Mathematics." 

Pascal  (1632-1662),  French.  A  great  mathematician  and  metaphy- 
sician. 

Plato  (429-348  b.c),  Greek.  A  pupil  of  Socrates  and  at  head  of  a 
school  in  Athens.  Many  of  Euclid's  definitions  are  ascribed  to  him.  He 
was  the  first  to  use  the  method  of  analysis  to  discover  proof  of  theorem. 
He  added  much  to  geometrical  knowledge. 

Poncelet,  J.  V.  (1788-1867),  French.  He  investigated  and  developed 
modern  projective  geoinetry. 

Ptolemaeus,  Claudius  (about  139  a.d.),  Egyptian.  A  celebrated  a.s- 
tronomer.  The  fundamental  idea  of  his  system  as  opposed  to  that  of 
Copernicus  was  that  the  earth  is  the  center  of  the  universe,  and  that  the 
sun  and  planets  revolve  about  it. 

Pythagoras  (580  b.c),  Greek,  but  lived  many  years  in  Egypt.  In 
lower  Italy  he  founded  a  school  for  the  teaching  of  mathematics,  philoso- 
phy and  the  natural  sciences,  chiefly  the  first. 

Steiner  (1796-1863),  Swiss-German.  "  The  greatest  geometrician  since 
Euclid."     He  laid  the  foundation  for  modern  synthetic  geometry. 

Thales  (640-546  b.c),  Greek.  Lived  long  in  Egypt.  He  measured 
the  height  of  the  Pyramids  from  their  shadows.  He  was  one  of  the  Seven 
Wise  Men  of  Greece. 


INDEX 


Algebra  and  geometry,  connected, 

120. 
Alternation,  proportion  by,  105. 
Altitude  of  cone,  306. 

cylinder,  282. 

frustum  of  pyramid,  288. 

prism,  267. 

prismoid,  288. 

pyramid,  287. 

rhomboid,  53. 

trapezoid,  53. 

triangle,  30. 
Analysis  of  problems,  12,  120. 
Angle,  4. 
acute,  5. 

between  any  two  circles,  328. 
between  two  curves,  111. 
central,  111. 
dihedral,  253. 
escribed.  111. 
exterior,  of  triangle,  30. 
initial  line  of,  4. 
inscribed,  111. 
negative,  5. 
oblique,  5. 
obtuse,  5. 
polyhedral,  254. 
positive,  5, 
rectilineal,  254. 
right,  4. 

same  kind  of,  6. 
spherical,  328. 
straight,  5. 
terminal  line  of,  4. 
tetrahedral,  255. 
trihedral,  255. 
vertex  of,  4. 


Angles,  adjacent,  definition  of,  6. 
adjacent  and  vertical,  Group  on, 

19-22. 
alternate  exterior,  6. 
alternate  interior,  6. 
classes  of, 

as  to  their  algebraic  sign,  4. 
as  to  their  size,  5. 
as  to  their  location,  6. 
complemental,  6. 
corresponding,  6. 
exterior,  6. 
interior,  6. 

measurement  of,  100-119. 
right,  Group  on,  32-34. 
supplemental,  6. 
vertical,  6,  19. 
Antecedent,  101. 
Apothem,  210. 
Arc,  78. 

Arc-radius,  313. 
Area,  definition  of,  138. 
of  the  circle,  Group  on,  220-224. 
unit  of,  138. 
Areal  measurement 
in  the  plane,  333. 
on  the  sphere,  333. 
ratios.  Group  on,  176-179. 
Areas  of  rectangles  and  other  poly- 
gons. Group  on,  138-148. 
Axioms,  9,  331. 
Axis  of  circle  of  a  sphere,  311. 
revolution,  283. 
the  surface,  283, 
radical,  208. 


Babylonians,  113,  Note.; 


J49 


350 


INDKX 


Base  of  cylinder,  282. 

prism,  266. 

projection,  149. 

pyramid,  287. 

triangle,  30. 
Biographical  sketches,  847,  348. 
Bisector,  definition  of,  1. 

Center,  circum-,  of  triangle,  94. 

ex-,  of  triangle,  03. 

in-,  of  triangle,  91. 

of  circle,  8. 

of  gravity,  96. 

of  parallelepiped,  286. 

of  regular  polygon,  210. 

of  similitude,  160. 

ortho-,  of  triangle,  95. 

radical,  of  three  circles,  209. 
Centroid  of  triangle,  96. 
Chart  problems,  77. 
Chord,  definition  of,  78. 
Circle,    area    of.    Group    on,   220- 
227. 

axis  of ,  311. 

center  of,  8. 

circumference  of,  8. 

great,  311,  331. 

small,  311. 
Circle  and  its  related  lines,  Group 

on,  79-90. 
Circles,  circumscribed,  138. 

concentric,  8. 

escribed,  93. 

inscribed,  93. 
Circumference,  8. 
"Circumscribed  figures,  138,  210-216, 

311. 
Coincident  superposition,  2,  47. 
Commensurable  ratios,  101 ,  102. 
Complemental  angles,  6. 
*        Complex  figure,  definition  of,  2. 
Composition,  proportion  by,  105. 
Composition  and  division,  propor- 
tion by,  106. 
Conclusion,  definition  of,  11. 


Concurrent  lines  of  triangle,  Group 
on,  91-99. 

transversals  and  normals,  Group 
on,  228-232. 
Cone,  altitude  of,  306. 

base  of,  306. 

circular,  306-308. 

definition  of,  306. 

elements  of,  306. 

frustum  of,  807. 

Group  on,  306-:^08. 

of  revolution,  307. 

right,  307. 

truncated,  307. 
Congruent  figures,  definition  of,  2. 

triangles,  Group  on,  45-48. 
Conical  surface,  definition  of,  306. 
directrix  of,  300. 
elements  of,  300. 
generatrix  of,  306. 
nappes  of,  306. 
vertex  of,  300. 
Conjugate  harmonic  points,  184. 
Consequent,  definition  of,  101. 
Constant,  definition  of,  108. 
Construction  of  a  figure,  12. 
Continued  proportion,  103. 
Contradiction,  definition  of,  11. 
Converse  of  a  theorem,  11. 
Convex  spherical  polygon,  329. 

surface  of  a  pyramid,  287. 
Corollary,  definition  of,  11. 
Corresponding  points,  160. 
Couplet,  definition  of,  160, 
Cube,  definition  of.  267. 
Cyclic  four-side,  53. 
Cylinder,  altitude  of,  282. 

bases  of,  282. 

circular,  282-284. 

definition  of,  282. 

elements  of,  282. 

Group  on,  266-282. 

of  revolution,  283. 

right,  282. 

right  section  of,  282. 


INDEX 


351 


Cylindrical  surface, 

definition  of,  282. 
directrix  of,  282. 
elements  of,  282. 
generatrix  of,  282. 

Decagon,  3. 
Definition,  1. 
Degree,  113. 

Determinate  problems,  130,  131. 
Determination  of  circle,  331. 
locus,  73. 
plane,  233,  234. 
Diagonal  of  polygon,  63. 
Diameter  of  circle,  78. 
Dihedral  angle, 

definition  of,  253. 

edge  of,  253. 

faces  of,  253. 

measure  of,  254. 

method  of  reading,  253. 

rectilineal  angle  of,  254. 

right,  253. 
Discussion  of  problem,  12. 
Distance  from  point  to  line,  38. 

on  surface  of  sphere,  328. 
Division,  harmonic,  184. 

proportion  by,  106. 
Dodecagon,  3. 
Dodecahedron,  266. 

Edge  of  dihedral  angle,  253. 
Edges  of  polyhedral  angle,  266. 

polyhedron,  254. 
Elements  of  cone,  306. 

conical  surface,  306. 

cylinder,  282. 

cylindrical  surface,  282. 
Equal  figures,  2. 

Escribed  circles,  93  ;  angles,  111. 
Ex-center  of  triangle,  93. 
Exterior  angle  of  triangle,  30. 
Extreme    and    mean     ratio,     196, 

197. 
Extremes,  definition  of,  103. 


Faces  of  dihedral  angle,  253. 

polyhedral  angle,  254. 
polyhedron,  266. 
Figures,  areas  of  irregular,  144. 
definitions  of^  2. 
similar.  Group  on,  160-175. 
Foot  of  a  line,  234. 
Fourth  proportional,  172,  194. 
Frustum  of  cone,  307,  308. 

pyramid,  288,  298,  299, 
301,  305. 

Golden  Section,  190. 

Harmonic  division,  184, 
Heptagon,  3. 
Hexagon,  3. 
Homologous  lines,  160. 
Hypotenuse,  30,  38. 
Hypothesis,  definition  of,  11. 

Icosahedron,  266. 
In-center  of  triangle,  91. 
Incommensurable  ratios,  102. 
Indeterminate  problems,  130,  131. 
Inscribed  polygon,  138. 
Inversion,  proportion  by,  105. 
Isoangular  triangle,  30,  32. 
Isosceles  triangle, 

definition  of,  30. 

part  of  sector  of  circle,  78. 
Isosceles  and  scalene  triangles,  Group 
on,  35-44. 

Join,  definition  of,  2. 

Lateral  area  of  prism,  267. 

edges  of  pyramid,  287. 
faces  of  pyramid,  287. 
surface  of  pyramid,  287. 
Limit,  definition  of,  108. 
Limits,  method  of,  108,  109. 

postulate  of,  109. 
Line,  broken,  2 ;  curved,  2. 
definition  of,  1.  '     . 


352 


INDEX 


Line,  on  sphere,  331. 

straight,  1. 
Linear    application    of   proportion, 

Group  on,  18;{-209. 
Lines,  concurrent,  2. 

homologous,  100. 

parallel,  7,  23,  234. 

perpendicular,  4,  23,  234. 
Lines  and  mid-joins.  Group  on, 01, 68. 
Line-segments,  1,  189. 

projection  of,  149. 
Locus,  definition  of,  8. 

determination  of,  73. 

exercises  on,  123-126. 

illustrations  of  elementary  princi- 
ples of,  74-77. 
Lune,  angle  of,  331. 

area  of,  331. 

definition  of,  331. 

Mean  proportional,  103. 
Means,  definition  of,  103. 
Measure,  angular,  100, 

radical,  101. 
Measurement,  Group  on,  100-110. 

of  angles,  Group  on,  111-137. 
Median  of  trapezoid,  53. 

triangle,  80. 
Method  of  limits,  108,  109. 
Mid-joins  and  lines.  Group  on,  01-08. 
Mid-normal,  or   mid-perpendicular, 
4,  09,  70. 

Nappes  of  conical  surface,  306. 
Nine-point  circle  theorem,  128. 
Normals,    concurrent,    Group    on, 

228-232. 
Numerical  measure,  100. 

Octagon,  3. 
Octahedron,  266. 
Opposite  of  theorem,  11. 
Orthocenter  of  triangle,  95. 
Orthogonal  circles,  theorem  of,  129. 
Orthogonally,  defined.  111. 


Parallel  group,  23-20. 

lines,  7,  23,  234. 

lines  to  a  plane,  245,  247. 

planes,  254. 
Parallelepiped,  definition  of,  207. 

division  of,  277. 

rectangular,  207,  272-276. 

volume  of,  270. 
Parallelogram,  definition  of,  63. 
Parallelograms,  areas  of,  141-143. 

Group  on,  53-t5(). 
Pedal  triangle,  127. 
Pentagon,  definition  of,  3. 
Perigon,  definition  of,  5. 
Perimeter,  computation  of,  222,  223. 

definition  of,  221. 
Perpendicular  lines,  4,  23. 

to  a  plane,  234. 
Perpendiculars,    theorems    relating 

to,  230-239,  241-243. 
Perspective,  160. 

PI  anal  angles,  Group  on,  253-205. 
Plane,  definition  of,  2,  233. 

figure,  2. 

geometry,  definition  of,  12. 

tangent  to  sphere,  311. 
Plane  and  its  related  lines.  Group 

on,  233-252. 
Point,  definition  of,  1. 
Points,    equidistant    and    random, 

Group  on,  09-77. 
Polar  distance  of  circle,  311. 

of  spherical  polygon,  330. 
Poles  of  a  circle,  311. 
Polygon,  angles  of,  32,  33. 

circumscribed,  138. 

definition  of,  2,  3. 

diagonal  of,  53. 

inscribed,  138. 

regular,  3,  170,  210. 

similar,  100,   166,  167,  170,   178, 
186,  206. 

spherical,  329,  330. 
Polygons,  area  of,  Group  on,   138- 
148. 


INDEX 


353 


Polygons,    circumscribed    and    in- 
scribed   regular,    Group    on, 
210-219. 
Polyhedra,  classified,  266. 

similar,  268. 
Polyhedral,  convex,  250. 

definition  of,  254. 

edges  of,  254. 

face  angles  of,  255. 

faces  of,  254. 

planal  angles  of,  254. 

vertex  of,  254. 
Polyhedrals,  method  of  reading,  255. 

symmetric,  255. 
Polyhedron,  convex,  266. 

definition  of,  266. 

edges  of,  266. 

faces  of,  266. 

sections  of,  266. 

vertices  of,  266. 

volume  of,  268. 
Postulate,  definition  of,  7. 

of  limits,  109. 
Principal  section,  definition  of,  323. 
Prism,  altitude  of,  267. 

bases  of,  266. 

definition  of,  266. 

Group  on,  282-284. 

lateral  area  of,  267. 

lateral  edges  of,  267. 

lateral  faces  of,  266. 

oblique,  267. 

pentagonal,  267. 

(quadrangular,  267. 

regular,  267. 

right,  267. 

right  section  of,  267. 

right  truncated,  267. 

triangular,  267. 

truncated,  267. 

volume  of,  279. 
Prismoid,  definition  of,  288. 
Prismoidal  formula,  303. 
Problem,  definition  of,  12. 

solution  of,  12,  120. 


Problems,  classification  of,  130, 131. 
Projection,  base  of,  149. 

of  line  on  plane,  234. 

of  line-segment,  149. 

of  point  on  line,  149. 

of  point  on  plane,  234. 
Proof,  definition  of,  11. 
Proportion,  103-107,  191. 

linear  application  of.  Group  on, 
183-209. 
Proportional,  construction  of,  194, 

195. 
Proposition,  definition  of,  12. 
Pyramid,  altitude  of,  287. 

base  of,  287. 

definition  of,  287. 

frustum  of,  288. 

lateral  faces  of,  287. 

lateral    (or   convex)    surface    of, 
287. 

quadrangular,  288. 

regular,  288. 

slant,  288. 

triangular,  288. 

truncated,  288. 

vertex  of,  287. 

volume  of,  297. 
Pythagorean  group,  149-159. 

Quadrilateral,  or  four-side,  53. 
cyclic,  53. 

Kadian,  definition  of,  187. 
Kadical  axes,  208. 

center,  209. 

plane,  335. 
Radius  of  circle,  8. 

regular  polygon,  210. 
sphere,  311. 
Ratio,  extreme  and  mean,  196,  197 

of  similitude,  160. 
Ratio  and  proportion,  101-104. 
Ratios,  areal,  Group  on,  176-179. 

commensurable,  101,  102. 

incommensurable,  103,  104. 


354 


INDEX 


Reciprocal  proportion,  191. 

theorem,  11. 
Rectangle,  definition  of,  53. 
Rectangles,  areas  of,  Group  on,  138- 

140. 
Relative  direction,  5. 
Rhomboid,  53. 
Right  angle,  definition  of,  4. 
Right  angles  group,  30-34. 
Ruled  surface,  243,  Ex.  11. 

Scalene  triangle,  definition  of,  30. 

Group  on,  38-41. 
Scholium,  12. 
Seciant,  78. 

Section,  principal,  323. 
Sector  of  circle,  78. 

spherical,  312. 
Segment  of  circle,  78. 
line,  1,  189. 

spherical,  312. 
Sides  of  an  angle,  4. 
Similar  figures, 

definition  of,  2. 
Group  on,  160-175. 
Similarity,  doctrine  of,  333. 
Similitude,  center  of,  160. 

ratio  of,  160. 
Solid,  definition  of,  1. 

geometry,  266. 
Solution  of  problem,  12,  120. 
Sphere,  area  of  surface  of,  316. 

areal  measurement  on,  333. 

axis  of  circle  of,  311. 

circumscribed  about,  311. 

definition  of,  311. 

diameter  of,  311. 

great  circle  of,  311. 

Group  on,  311-327. 

inscribed  in  polygon,  311. 

plane  tangent  to,  311. 

polar  distance  of,  311. 

poles  of,  311. 

radius  of,  311. 

small  circle  of,  311. 


Sphere   surface.   Group    on  geom- 
etry of,  328-346. 

volume  of,  318,  319. 
Spherical  angle,  328. 

excess,  331. 

geometry.  Group  on,  328-346. 

polygon,  329. 

sector,  312. 

surface,  311. 

triangle,  329. 

wedge,  331. 
S(iuare,  definition  of,  53. 
Square  roots  of  numbers,  219. 
Straight  line,  1,  2  ;  angle,  5. 
Sum  of  angles,  5,  19. 
Superposition,  coincident,  2. 
Supplemental  angles,  6. 
Surface,  conical,  306. 

cylindrical,  282. 

definition  of,  1. 

spherical,  311. 

Tangent  to  a  circle,  78. 

plane  to  sphere,  311. 
Tetrahedrals,  255. 
Tetrahedron,  266. 
Theorem,  definition  of,  11. 
Theorems  of  special  interest,  127-130. 

on  inequality,  10. 
Third  proportional,  103,  195. 
Transformation,  definition  of,  103. 

of  figures,  199,  203. 
Transversal,  definition  of,  2. 

plane,  254. 
Transversals,     concurrent,     Group 

on,  228,  229. 
Trapezium,  definition  of,  53. 
Trapezoid,  altitude  of,  53. 

area  of,  144. 

definition  of,  53. 

isosceles,  53. 

median  of,  63. 

mid-join  of,  53. 
Triangle,  acute,  30. 

altitude  of,  30. 


INDEX 


355 


Triangle,  area  of,  143. 

base  of,  30. 

concurrent   lines  of,   Grpup    on, 
91-99. 

definition  of,  3. 

equiangular,  30. 

equilateral,  30. 

exterior  angle  of,  30. 

hypotenuse  of,  30. 

isoangular,  30. 

isosceles.  30,  35-37. 

median  of,  30,  m. 

obtuse,  30. 

right,  30. 

scalene,  30,  38-41. 

spherical,  329. 

vertex  angle  of,  30. 
Triangles,  congruent.  Group  on,  45- 
52. 

similar,  160,  103-165. 
Triangular  relations,  summaiy  of, 
98,  99. 


Trihedral,  birectangular,  265. 
definition  of,  255. 
rectangular,  255. 
trirectangular,  256. 

Ungula,  331. 
Unit  of  area,  138. 

length,  100. 

surface  (sphere),  333. 

volume,  268. 

Variable,  108. 
Vertex  of  angle,  4. 

conical  surface,  306. 
polyhedral,  254. 
pyramid,  287. 
angle  of  triangle,  30. 
Vertical  angles,  Group  on,  19-22. 
Volume  of  polyhedron,  268. 
unit  of,  268. 

Wedge  (spherical),  331. 

Zon*,  311. 


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